Certain one-relator products of groups: Freiheitssatz and non-triviality
Ihechukwu Chinyere
Submitted for the degree of Doctor of Philosophy
Heriot-Watt University
Department of Mathematics, School of Mathematical and Computer Sciences.
November 6, 2015
The copyright in this thesis is owned by the author. Any quotation from the thesis or use of any of the information contained in it must acknowledge this thesis as the source of the quotation or information. Abstract
Combinatorial group theory is an aspect of group theory that deals with groups given by presentations. Many techniques both geometric and algebraic are used in tackling problems in this area. In this work, we mostly adopt the geometric approach. In particular we use pictures and clique-pictures to describe various properties of certain one-relator product of groups. These groups are our object of study and they have presentations of the form G G = Λ , N(w) where GΛ is a free product of groups Gλ∈Λ and w is a (cyclically reduced) word in the free product GΛ with length at least two. It is known that most results about G are obtained by putting some conditions on the Gλ’s, on w, or both. In the case where Λ = 2, say Λ = {1, 2}, there are two forms of conditions we put on w. The first condition is that w = rn, where r is cyclically reduced word of length at most eight and n > 0. More precisely suppose A and B non-trivial groups and G = (A ∗ B)/N(rn). Suppose that any of the following conditions holds:
1. `(r) = 8 and n ≥ 3; or
2.2 ≤ `(r) < 8 and n ≥ 2; or
3. n = 1 and `(r) = 4 say r = abcd, with ha, ci and hb, di isomorphic to 2-
generator subgroups of PSL2(C).
We show that at least one of the following holds:
i. A and B embed in G naturally and r has order n in G.
ii. n = 2 and up to conjugation r has the form r = axbx−1cz with z2 = 1, where a, b, c ∈ A and x, z ∈ B. We then deduce that r has order n in G. The other condition is that w = rn as before but we require also that r is a word in −1 hai ∗ hUbU i, where a, b ∈ G1 ∪ G2 and U is a word in G1 ∗ G2. We prove that a minimal clique-picture over G satisfies the small-cancellation condition C(6) under the condition that r has no element of order two or r has length at least two in the −1 generators {a, UbU } and whenever both a and b belong to same factor, say G1, then either the subgroup of G1 generated by {a, b} is cyclic or hai ∩ hbi = 1. From this we deduce the following results.
−1 n 1. Let H be the quotient of hai ∗ hUbU i by the normal closure of R . Then G1,
G2 and H all embed in G via the natural maps.
2. If the word problems are soluble for H, G1 and G2, then it is soluble for G.
3. No proper cyclic sub-word of w is trivial in G. In particular r has order n in G.
We also consider the case where Λ = 3, say Λ = {1, 2, 3}, and prove the following results.
1. Suppose G1, G2 and G3 are all finite cyclic groups, then each embed in G via the natural maps provided w has non-zero exponent sum in each of the generators.
2. If w has length at most eight, then G is non-trivial.
1 Acknowledgements
I am most grateful to God Almighty for giving me life, good health and peace of mind throughout the period of this work. My deepest appreciation goes to my Supervisor Prof. Jim Howie who introduced me to combinatorial group theory and carried me along with his ideas and suggestions. I thank Prof. Bernd Schroers and Dr. Emma Coutts through whom I got to know about Jim and Heriot-Watt in general. Many thanks to Prof. Ingrid Rewitzky, Dr. Cornelia Naude and Dr. Bruce Bartlett for their assistance while I was at Stellenbosch University. Many thanks to my examiners Prof. Nick Gilbert and Dr. Martin Edjvet for the wonderful job done in reading my work. I also thank the entire HWU MACS staff who have worked together to make my stay here comfortable and successful. I thank my office mates and friends for their warm company. Special thanks to Bernard Oduoku for helping in the proof-reading. This work would not have been possible without the Maxwell Scholarship. I thank Heriot-Watt University for granting me this life time opportunity. Finally I thank my family for their love, prayer and support. It is you who have had to put up with my long period of absence. I hold you all in high esteem and would never let you down. Contents
Introduction ii 0.1 Layout ...... 3
1 Preliminaries 5 1.1 Preamble ...... 5 1.2 Free groups and Group presentations ...... 5 1.3 Algorithmic problems ...... 6 1.3.1 Tietze transformations ...... 7 1.3.2 Nielsen Equivalence ...... 7 1.4 Free Product, Free Product with Amalgamation and HNN extension .8 1.4.1 Free Products ...... 8 1.4.2 Free Product with Amalgamation and HNN extensions . . . .9 1.5 Small cancellation ...... 11 1.5.1 Basic formulas ...... 12 1.5.2 Pictures ...... 13 1.5.3 Combinatorial curvature ...... 16 1.6 Bass-Serre Theory ...... 17 1.6.1 Group actions ...... 17 1.6.2 Group actions on graphs ...... 18 1.6.3 Theory of covering spaces ...... 19 1.6.4 Graph of groups ...... 19 1.6.5 The fundamental group of a graph of groups ...... 20 1.6.6 Group action on trees ...... 21
2 Short review of one-relator product of groups 24 2.1 Preamble ...... 24 2.2 One relator groups ...... 24 2.3 One relator product of groups ...... 25 2.4 One relator product of cyclics ...... 27 2.5 Relative presentations ...... 29
i CONTENTS
3 One-relator product induced from generalised triangle group 31 3.1 Preamble ...... 31 3.2 Periodicity ...... 33 3.3 Generalised triangle group description and refinements ...... 35 3.4 Clique-pictures ...... 37 3.5 Clique-labels and virtual periodicity ...... 42 3.6 The relator has free product length at least 4 and the pair (a, b) is admissible...... 47 3.7 The relator has no letter of order 2 ...... 49 3.7.1 Proof of Theorem 3.7.1 ...... 58 3.8 Proof of Theorems ...... 62
4 One-relator product of two non-trivial groups with short relator 66 4.1 Preamble ...... 66 4.2 Preliminary results ...... 68 4.3 Relator of length eight with power n ≥ 3...... 73 4.4 Relator of length six with power n ≥ 2 ...... 79 4.4.1 Configurations of an interior vertex of degree 4 ...... 82 4.4.2 Configurations of an interior vertex of degree 5 ...... 88 4.5 Relator of length four with power n ≥ 2 ...... 99 4.6 Special case of n = 1 and relator has length four...... 103
5 One-relator product of three non-trivial finite cyclic groups 106 5.1 Preamble ...... 106 5.2 S1-equivariant homotopy and degree of maps ...... 106 5.3 Main result ...... 109
6 One-relator product of three non-trivial groups with short relator112 6.1 Preamble ...... 112 6.2 Technical Lemmas ...... 113 6.3 Main result ...... 123
7 Conclusion and future work 124 7.0.1 Removing admissibility condition ...... 124 7.0.2 Solution of equations over free product of groups ...... 125
ii List of Tables
4.1 A table containing all the possible configurations of a positively- curved interior vertex...... 74 4.2 A table containing all the possible configurations of a positively- curved interior vertex...... 82
6.1 A table containing all the possible words R (up to cyclic permutation
and inversion) in {U2,V } of index k, 1 ≤ k ≤ 3 and `(R) ≥ 2(6 − k). 119
iii List of Figures
1.1 Diagram showing bridge-move...... 14 1.2 A graph with two edges and one vertex...... 22 1.3 Graph of groups for free product of A and B...... 22 1.4 Graph of groups for free product of A and B amalgamated over C.. 23 1.5 Graph of groups for HNN extension...... 23
3.1 Push-out diagram...... 31 3.2 Double push-out diagram...... 36 3.3 Diagram showing an example of u ∼ v...... 38 3.4 Diagram showing u and v joined by l/2 arcs after doing bridge-moves to Figure 3.3...... 38 3.5 Diagram showing the l/2-zone in Figure 3.4 replaced by a single shaded rectangle...... 39 3.6 Diagram showing two vertices u ∼ v joined by l-zone...... 39 3.7 Diagram showing the l-zone in Figure 3.6 replaced by a single shaded rectangle...... 39 3.8 Diagram showing a simply-connected clique consisting of only two vertices and how the two vertices are combined to form one single vertex...... 40 3.9 Diagram showing a non-simply connected clique...... 40
3.10 Diagram showing vertices v1 and v5 joined by a 2-zone and vertices
v2 and v5 joined by a 3-zone...... 41 3.11 Diagram showing a vertex v of degree five...... 43
3.12 Diagram showing the case where only the special letters z0, zl/2 and
zl are contained in zones ...... 52
3.13 Diagram showing the case where all four special letters z0, zl/2, zl and
z3l/2 are contained in zones ...... 53
3.14 Diagram showing a clique containing vertices v1, v2, . . . , v5 which is
not simply-connected. The region bounded by v1, v5, v4, v5 and con-
taining v6 (indicated by the red curve) contains only simply-connected
cliques, and so is a good choice for C since v6 is in P ∩ C...... 63
iv LIST OF FIGURES
4.1 A picture consisting of only k boundary vertices joined to each other by l-zones...... 71 4.2 Diagram showing adjacent 6-zones of a configuration of Type 1 . . . . 75 4.3 Diagram showing configuration (6, 3, 6, ∗, ∗) in Type 1 ...... 75 4.4 Diagram showing configuration (6, 4, 6, ∗, ∗) in Type 2 ...... 76 4.5 Diagram showing configuration (6, 5, 6, ∗, ∗) in Type 2 ...... 76 4.6 Diagram showing configuration (6, 4, 5, ∗, ∗) in Type 3 ...... 77 4.7 Diagram showing configuration (6, 5, 5, ∗, ∗) in Type 3 ...... 77 4.8 Diagram showing configuration (6, 5, 4, 4, 5) in Type 4 ...... 78 4.9 Configuration (4, 4, ∗, ∗, ∗)...... 79 4.10 Diagram showing positively-curved boundary vertex of degree 3 . . . 81 4.11 Configuration (4, 4, 3, 1) ...... 83 4.12 Configuration (4, 3, 4, 1) ...... 84 4.13 Configuration (4, 4, 2, 2) ...... 84 4.14 Configuration (4, 3, 3, 2) ...... 85 4.15 Configuration (4, 3, 2, 3) ...... 86 4.16 Configuration (3, 3, 3, 3) ...... 87 4.17 Configuration of Type 2 ...... 88 4.18 Adjacent 4-zones ...... 88 4.19 Two 4-zones separated by a 1-zone ...... 89 4.20 Configuration (4, 1, 3, 3, 1) ...... 89 4.21 Configuration (4, 1, 2, 3, 2) ...... 90 4.22 Configuration (4, 1, 3, 2, 2) ...... 91 4.23 Configuration (4, 2, 3, 1, 2) ...... 92 4.24 Configuration (3, 3, 2, 2, 2) ...... 93 4.25 Configuration (3, 2, 3, 2, 2) ...... 94 4.26 Configuration (3, 3, 3, 2, 1) ...... 95 4.27 Configuration (3, 3, 1, 3, 2) ...... 96 4.28 Configuration (4, 2, 2, 2, 2) ...... 97 4.29 Configuration (4, 2, 4, 1, 1) ...... 98 4.30 Configuration (4, 2, 4, 2) ...... 99 4.31 A segment T of X ...... 100 4.32 ...... 101 4.33 A positively-curved interior vertex ...... 102
6.1 Diagram showing a section of the tree Γ on which H acts...... 114 6.2 Diagram showing an (A ∗ B)-region with label (U 2V )2UV of index 2. The head of the red arrows indicate the adjacent C-regions receiving π/3 curvature...... 120
v LIST OF FIGURES
6.3 Positively oriented vertices of Γ when n = 2. The figure on the left 2 n corresponds to a vertex of Γ when r = (t c1tc2) and the other is −2 n when r = (t c1tc2) ...... 121
vi Introduction
For some indexing set Λ, let GΛ be the free product of groups Gλ, where λ ∈ Λ. The object of study in this thesis falls under the class of groups of the form
G G = Λ , N(w)
otherwise known as a one-relator product of groups Gλ, with relator w. Such groups have been widely studied for various reasons. One of such reasons follows from the fact that G generalises one-relator groups. Hence one can try to see how much the rich theory of one-relator groups can be generalised. On the other hand many problems in combinatorial group theory and low dimen- sional topology are indeed problems about one-relator products. For us the moti- vation to study one-relator products comes from a simple construction in topology known as Dehn surgery. As the name implies, this construction is due to Max Dehn in 1910, motivated by the works of French mathematician Henri Poincar´e. Poincar´eat the end of a long article [80] asked the following seemingly innocuous question:
Is every closed 3-manifold with trivial fundamental group topologi- cally equivalent to the 3-dimensional sphere S3?
An affirmative answer to the above question was known as Poincar´econjecture, now a theorem of Perelman (see [11]). For us the interesting bit is that this was not the first formulation of the problem. The earlier version was in terms of homology rather than fundamental group, of which Poincar´efound a counter-example himself. Dehn [17] then invented the Dehn surgery construction to show that there are in fact an infinite family of homology spheres which are all counter-examples to the earlier version of this problem. Dehn surgery on a knot K in S3 is the following. Remove a regular neighbourhood N(K) of K from S3 to get the manifold S3(K) ∼= S3 − N(K); glue in a solid 3 ∼ torus using some boundary homeomorphism to get the surgered manifold Ss (K) = 3 1 2 S (K) ∪s (S × D ), where s ∈ Q ∪ {∞}, also known as surgery slope, is the isotopy
1 LIST OF FIGURES class of essential simple closed curves on the boundary of N(K). The surgered 3 manifold Ss (K) determined up to homeomorphism by s. The knot K can be replaced by a link, in which case the surgery is done on each component. In this setting, an important result due to Lickorish and Wallace [[68], [94]] states that every closed, orientable, connected 3-manifold is obtained by Dehn surgery. There have been hundreds of publications on this subject alone. To learn more about history and scope, we refer the reader to [35]. If we restrict to the case where K is a knot, a basic question to ask is what the 3 3 possibilities for Ss (K) can be. Even more basic is the question of when Ss (K) is reducible (i.e. decomposes as non-trivial connected sum or homeomorphic to S1 ×S2 (in the orientable case)). The first non-trivial examples of this phenomenon were provided by Moser, who showed in [78] that
3 ∼ Spq(Tp,q) = L(p, q)#L(q, p), where Tp,q is a (p, q)-torus knot and L(p, q),L(q, p) are lens spaces. Gabai showed in [30] that S1 ×S2 occurs only when K is the unknot U and s = 0. It follows that any 3 non-trivial reducible Ss (K) decomposes as a connected sum. It is also known that 3 Ss (K) can be reducible if K is a cabled knot [36]. If Cp,q with p > 1, is a (p, q)-cable of a knot C, then 3 ∼ 3 Spq(Cp,q) = L(p, q)#Sq/p(C). These are the only known examples where the phenomenon occurs. Of course one can think of Tp,q as a (p, q)-cable of U. This motivates the following conjecture due to Gonzalez-Acuna and Short [34].
3 3 Conjecture 0.0.1. (Cabling Conjecture) If K is a non-trivial knot in S and Ss (K) is reducible, then K = Cp,q and s = pq. A lot is known in support of an affirmative answer to this conjecture. For example it known that the conjecture is true for symmetric knots [[47], [46], [25]], knots with low bridge-number [[42], [48], [85], [59], [97]], alternating knots [77], arborescent knots [96], satellite knots [86] and strongly invertible knots [37].
3 Also it is known that if Ss (K) is reducible, then s must be an integer [[38], [7], see 3 3 also [34]], and Ss (K) contains a lens space summand. Moreover when Ss (K) is the connected sum of two lens spaces, then K is either a torus knot or a cable of a torus knot [40]. Other results by Howie [58] and Valdez S´anchez [93] imply that three is the upper bound for the number of summands, and if three summands occur, two must be lens spaces and the third is an integer homology sphere. Conjecture 0.0.2. (Two-summand Conjecture) If K is a non-trivial knot in S3 3 3 ∼ and Ss (K) is reducible, then Ss (K) = M1#M2, with M1 and M2 irreducible.
2 LIST OF FIGURES
Clearly, the Cabling conjecture implies the Two-summand conjecture. The latter is known to be true for slice knots [76], knots with bridge number at most five and knots with positive braid closures [97]. In particular both conjectures lead to questions about one-relator products of groups with two or three factors. Since knot groups have single weight, a question of when such groups are trivial arises. This thesis studies these groups under various conditions on the factors, or the relator, or both.
0.1 Layout
There are seven chapters: Chapters 1, 2 and 7 are the three minor chapters while Chapters 3, 4, 5 and 6 are the major chapters. In Chapter 1, we mention a few basic concepts which will be used throughout this thesis. In particular we define pictures which is our main geometric tool for proving various results. Chapter 2 contains a brief review of groups with one-relator presentations. We mention some important existing results about them.
n In Chapter 3, we consider the one-relator group G = (G1 ∗ G2)/N(R ), where R is −1 a word in the free product hai ∗ hUbU i for some letters a, b ∈ G1 ∪ G2 and word
U ∈ G1 ∗ G2. We show that if n > 1 and R has no letter of order 2, or `(R) > 2 as a cyclically reduced word in the free product hai ∗ hUbU −1i and (a, b) is admissible, then the following holds.
1. The natural maps G1 → G, G2 → G and H → G are all injective, where H is the quotient of hai ∗ hUbU −1i by N(Rn).
2. If the word problems are soluble for H, G1 and G2, then it is soluble for G.
3. No proper cyclic subword of Rn is trivial in G. In particular R has order n in G.
We do this using (clique-)pictures which we shall also describe in this chapter. In Chapter 4, we consider the one-relator product G = (A ∗ B)/N(rn) under the following conditions:
1. `(r) = 8 and n ≥ 3; or
2.2 ≤ `(r) < 8 and n ≥ 2; or
3. n = 1 and `(r) = 4 say r = abcd, with ha, ci and hb, di isomorphic to 2-
generator subgroups of PSL2(C).
3 LIST OF FIGURES
We show that either:
i. A and B embed in G naturally and r has order n in G, or
ii. n = 2 and up to conjugation r has the form r = axbx−1cz with z2 = 1, where a, b, c ∈ A and x, z ∈ B.
We then deduce that r has order n in G.
In Chapter 5, we consider a one-relator product G of three finite cyclic groups Ga,
Gb and Gc with generators a, b, and c respectively, and relator w. We show that if the exponent sum of each of the three generators in w is non-zero (modulo their respective orders), then the natural maps Ga → G, Gb → G and Gc → G are all injective. In Chapter 6, we apply various techniques and results from previous chapters to show that a one-relator product of three non-trivial groups is non-trivial under the condition that the relator has free product length at most 8. Chapter 7 is the concluding chapter. It contains the summary of this work as well various open problems related or arising from this work.
4 Chapter 1
Preliminaries
1.1 Preamble
In this Chapter we define some basic concepts that are relevant in this thesis. More specific definitions shall be made in the Chapters where they are first used. We begin with the concept of free groups and group presentations.
1.2 Free groups and Group presentations
Combinatorial group theory is mostly about group presentations. We mention a few things about free groups and group presentations. Our account is based on [65]. Definition 1.2.1. A group F (X) is said to be free on X ⊆ F (X) if, given any group G and any map φ : X → G, there is a unique homomorphism φ0 : F (X) → G extending φ. We call X the basis and |X| the rank of F (X). Details about existence and properties of F (X) can be found in [65]. By the above definition, we know that given any group G and a subset X of G such that every element in G is a word in X ∪ X−1, the inclusion map φ of X into G extends to a homomorphism φ0 from F (X) onto G. Hence it follows from the First isomorphism theorem that
F (X) G ' , (1.1) K where K is the kernel of φ0. This tells us that every group is the quotient of some free group. It also leads to the idea of group presentation. Definition 1.2.2. Let X be a set and R, a subset of F (X). A (free) presentation of a group G is a pair ℘ = hX | Ri, such that
F (X) G = , (1.2) N(R)
5 Chapter 1: Preliminaries where N(R) is the normal closure of R in F (X). Henceforth, we shall use G instead of ℘.
In Definition 1.2.1, the elements of X and R are called the generators and relators of G respectively. We say that G is finitely generated if |X| < ∞. If in addition |R| < ∞, we say G is finitely presented. The following proposition tells us when a mapping between two groups can be extended to a group homomorphism.
Proposition 1.2.3. Given groups G = hX | Ri and H = hY | Si, a mapping φ : X −→ H extends to a homomorphism φ0 : G −→ H if for all x ∈ X and r ∈ R, the result of substituting φ(x) for x in r yields a word in N(S).
1.3 Algorithmic problems
Although presentations are nice, sometimes it is hard to tell some basic properties of a group by presentation. In 1912 Max Dehn listed three problems about finitely presented groups. Let G be a group given by a finite presentation
G = hX | Ri. (1.3)
1. (Word problem) Is there an algorithm which decides whether or not any given word in G is the identity?
2. (Conjugacy problem) Is there an algorithm which decides whether or not any pair of words in G are conjugate?
3. (Isomorphism problem) Is there an algorithm which decides whether or not any pair of finite presentations define isomorphic groups?
For this thesis, we shall focus only on the word problem. The works of Novikov, 1952, 1955, and, independently, Boone, and Britton, both in 1958, proved that the word problem has no general solution. In particular they produced a finitely presented group in which there is no effective way of showing that a word in the given generators is trivial or not. However, the word problem is known to have a solution for certain classes of groups. These includes free groups [[73], Corollary 1.2.2], one-relator groups [[73], Theorem 4.14], automatic groups [[24], Theorem 2.3.10], small cancellation groups [[39],[70],[92]], residually finite groups and word hyperbolic groups. In particular for word hyperbolic groups, the algorithm that solves the word problem is extremely efficient.
6 Chapter 1: Preliminaries
1.3.1 Tietze transformations
It easy to see that any group has infinitely many different presentations. In 1906, Tietze gave a way to move from one presentation to another of the same group. Let G be a group given by a finite presentation
G = hX | Ri. (1.4)
(T1) Addition of a relation: If r = 1 in G, then replace R with R ∪ {r}.
(T2) Removal of a relation: If r ∈ R and N(R − {r}) = N(R), then replace R with R − {r}.
(T3) Addition of a new generator: Replace X with X ∪ {t} and R by R ∪ {tw−1} for some t 6∈ X and word w in F (X).
(T4) Removal of a generator: Replace X with X − {x} and R by R − {r} where r = xw−1 if w and all other words in R do not contain x±1.
The transformations T1-T4 are called Tietze transformations. Tietze showed that any two presentations of the same group are connected by repeated application of Tietze transformations.
1.3.2 Nielsen Equivalence
Let F be a free group on X and U a finite subset of F . We think of U as an n-tuple
(u1, u2, . . . , un) of elements of F .
Definition 1.3.1. An elementary Nielsen transformation of U is one of the following three types: for some i, 1 ≤ i ≤ n,
(T0) delete ui if ui = 1,
−1 (T1) replace ui by ui ,
(T2) replace ui by uiuj, j 6= i and leave uk fixed for all k 6= i. A Nielson transformation is a finite sequence of elementary Nielsen transformations.
Definition 1.3.2. Two subsets U and V of F are called Nielsen equivalent if U is carried into V by some Nielsen transformation τ. In such a case we write V = Uτ .
Lemma 1.3.3. The subgroups of F generated by U and Uτ are equal, for any Nielsen transformation τ.
7 Chapter 1: Preliminaries
A generating set U = {u1, u2, . . . , un} is called Nielsen reduced if the following conditions are satisfied for every ui, uj and uk in U.
1.1 6∈ {ui, uj, uk}
2. uiuj 6= 1 implies `(uiuj) ≥ `(ui), `(uj)
3. uiuj 6= 1 and ujuk 6= 1 implies `(uiujuk) > `(ui) − `(uj) + `(uk) The point here is that Nielsen reduced sets are free generators for the subgroup they generate.
1.4 Free Product, Free Product with Amalgama- tion and HNN extension
In this section we study the definitions, some properties and theorems of free prod- ucts, free product with amalgamation and HNN extensions. The reference for this section is [70]. We begin with free products.
1.4.1 Free Products
Definition 1.4.1. Let A1,A2,...,An be non-trivial groups for some n ∈ N. Suppose that Aλ = hXi | Rii and where Xi ∩ Xj = ∅ for i 6= j. The free product, ∗Ai =
A1 ∗ A2 ∗ ... ∗ An, of groups Ai is the group
n n ∗Ai = h∪i=1Xi | ∪i=1 Rii. (1.5)
The groups Ai are called the factors of ∗Ai. It is known that free product is inde- pendent of the choice of presentations for Ai. Definition 1.4.2. (Normal Form) Let G be a group with trivial pairwise intersecting subgroups Aλ∈Λ, for an indexing set Λ. A reduced sequence (or normal form) is a sequence g1, g2, . . . , gk, with k ≥ 0, of elements of ∪λ∈ΛAλ such that each gi is in one of Aλ, and if i > 1 and gi ∈ Aλ then gi−1, gi+1 6∈ Aλ. Definition 1.4.3. (Reduced Word) A reduced word in a group G is a word of the form g1g2 . . . gk, for some reduced sequence g1, g2, . . . , gk in G. A word is said to be cyclically reduced
Note that it follows from Definition 1.4.2 that in a reduced word w = g1g2 . . . gk with k > 1, no gi is trivial.
Definition 1.4.4. (Cyclically Reduced Word) A word w = g1g2 . . . gk a group G said to be cyclically reduced if it is reduced and g1, gk do not belong to same Aλ (using the notation in Definition 1.4.2).
8 Chapter 1: Preliminaries
Theorem 1.4.5. (Normal Form Theorem for Free Products) In a free product A∗B, the following statements are equivalent.
1. If w = g1g2 . . . gk where k > 0 and g1, g2, . . . , gk is a reduced sequence, then w 6= 1 in A ∗ B.
2. Each w ∈ A ∗ B can be uniquely expressed as a product w = g1g2 . . . gk where
g1, g2, . . . , gk is a reduced sequence.
The Normal Form Theorem allows us to have a notion of length for elements in a free product. It also follows from the Normal Form Theorem that the word and conjugacy prob- lems are solvable for A ∗ B if the factors are finitely generated and these problems are solvable in them. Also elements of finite order in A ∗ B are conjugates of finite order elements in the factors.
Definition 1.4.6. (Length) A word w in has length k if its normal form is of the form g1g2 . . . gk. We denote the length of w by `(w).
Lemma 1.4.7. Let A and B be subgroups of a group G such that A ∪ B generates G, A ∩ B = {1}, and no reduced word in G is equal to 1. Then G = A ∗ B.
Next we state two of the most important theorems about free products. The first is due to Grushko (1940) and Neumann (1943), while the second is due to Kurosh (1934).
Theorem 1.4.8. Let F be a free group, and let φ : F → ∗Ai be a homomorphism of F onto ∗Ai. Then there is a factorisation of F as a free product, ∗Fi, such that
φ(Fi) = Ai
However most applications of Theorem 1.4.8 uses the following corollary of it.
Corollary 1.4.9. If G = A1 ∗ A2 ∗ ... ∗ Ak and the minimum number of generators of Ai is ni, then the minimum number of generators for G is n1 + n2 + ... + nk.
Theorem 1.4.10. Let G = ∗Ai, and let H be a subgroup of G. Then H is a free product, H = F ∗ (∗Hj) where F is a free group and each Hj is the intersection of
H with a conjugate of some factor Ai of G.
1.4.2 Free Product with Amalgamation and HNN exten- sions
Free product with amalgamation was introduced by Schreier (1926), while HNN extensions was a joint work G, Higman, B. H. Neumann and N. Neumann in 1949.
9 Chapter 1: Preliminaries
These two constructions have the same basic idea. Both involves two subgroups and an isomorphism between them. To describe these constructions, let G = hX | Ri and H = hY | Si be groups with subgroups A ⊆ G and B ⊆ H, such that there exists an isomorphism φ : A → B.
Definition 1.4.11. The free product of G and H, amalgamating the subgroups A and B is the group
G ∗A=B H = hX,Y | R, S, a = φ(a), a ∈ Ai. (1.6)
When A and B are the trivial groups, we get the free product of G and H. Another −1 way to think of G ∗A=B H is as the quotient of G ∗ H by N({aφ (a): a ∈ A}).
When the isomorphism between A and B is known, the shorter notation G ∗A H is preferred to G ∗A=B H.
Definition 1.4.12. A sequence c1, c2, . . . , ck, where k ≥ 0, of elements of G ∗ H will be called reduced if:
1. Each ci is in one of the factors G or H.
2. Successive ci, ci+1 come from different factors.
3. If k > 1, then no ci is in A or B.
4. If k = 1, then c1 6= 1.
Theorem 1.4.13. (Normal Form Theorem for Free Products with Amalgamation)
If c1, c2, . . . , ck, where k ≥ 1 is a reduced sequence, then c1c2 . . . ck 6= 1 in G ∗A H.
In particular, G and H are embedded in G ∗A H by the maps g −→ g and h −→ h.
Theorem 1.4.14. (Torsion Theorem for Free Products with Amalgamation) Every element of finite order in G ∗A H is a conjugate of a finite order element in G or H.
Theorem 1.4.15. (Conjugacy Theorem for Free Products with Amalgamation) Let k > 1 and u = c1c2 . . . ck a cyclically reduced word. Then every cyclically reduced conjugate of u is a cyclic permutation of u conjugated by an element of A
We move on to describe HNN extensions. In this case we want A and B above to be subgroups of G. Throughout g with or without subscript refers to an element of G. Also with or with subscript is either 1 or −1.
Definition 1.4.16. The HNN extension of G relative to A, B and φ is the group
G∗ = hX ∪ {t} | R, t−1at = φ(a), a ∈ Ai. (1.7)
10 Chapter 1: Preliminaries
We call G the base of G∗, t is the stable letter, and A and B are the associated subgroups.
1 2 k Definition 1.4.17. A sequence g0, t , g1, t , . . . t , gk, where k ≥ 0, is said to be −1 −1 reduced if there is no consecutive subsequence t , gi, t with gi ∈ A or t, gj, t with gj ∈ B.
1 2 k Lemma 1.4.18. (Britton’s Lemma) If a sequence g0, t , g1, t , . . . , gk, t is reduced
1 2 k ∗ and k ≥ 1, then g0t g1t . . . t gk 6= 1 in G .
1 2 Definition 1.4.19. A normal form is a sequence g0, t , g1, t , . . . , gk, where k ≥ 0, such that:
1. g0 is an arbitrary element of G,
2. if i = −1, then gi is a representative of a coset of A in G,
3. if i = 1, then gi is a representative of a coset of B in G, and
4. there is no consecutive subsequence t, 1, t−.
Theorem 1.4.20. (Normal Form Theorem for HNN extensions) Every element w of
∗ 1 2 1 2 G has a unique representation w = g0t g1t . . . gk, where w = g0, t , g1, t , . . . , gk is a normal form. In particular G embeds in G∗ via the map g 7→ g.
1.5 Small cancellation
In this section we recall some basic results about general maps. Almost every piece of information in this section is gotten from [70].
Let S be a subset of the Euclidean plane E2. We use ∂S to mean the boundary of S, S¯ denotes the topological closure of S, and −S will denote E2 − S.A vertex is a point in E2. An edge is a subset of E2 homeomorphic to the open unit interval. A region is a bounded subset of E2 homeomorphic to the open unit disk.
Definition 1.5.1. A map M is a finite collection of vertices, edges, and regions which are pairwise disjoint and satisfy:
1. If e is an edge of M, there are vertices u and v (possibly the same one) in M such thate ¯ = e ∪ u ∪ v.
2. The boundary, ∂D, of each region D of M is connected and there is a set of
edges e1, . . . , en in M such that ∂D =e ¯1 ∪ ... ∪ e¯n.
11 Chapter 1: Preliminaries
1.5.1 Basic formulas
In [69], Lyndon obtained results on maps which generalize the properties of regular tessellations of the plane. Previously Blanc [5] had used similar results but in a different context. The basic idea is the following. Let p and q be positive real numbers satisfying 1/p + 1/q = 1/2. The only integral solutions are the tuples (3, 6), (4, 4) and (6, 3). These pairs correspond to the tilings of the plane by regular triangles, squares, and hexagons respectively.
Some definitions
Let M be a map. A boundary vertex or boundary edge of M is a vertex or edge in ∂M.A boundary region is a region D of M such that ∂D ∩ ∂M 6= ∅. Note that by definition, if D is a boundary region of M, ∂D ∩ ∂M need not contain an edge. A vertex, edge or region which is not a boundary vertex, edge, or region is called interior. The degree of a vertex v, denoted d(v), is the number of edges incident on it. The degree of a region D, denoted d(D), is the number of edges on ∂D. The symbol i(D) will denote the number of interior edges of ∂D, with an edge counted twice if it appears twice in the boundary cycle of D. Definition 1.5.2. A non-empty map is called a [p, q]-map if each interior vertex has degree at least p and all regions have degree at least q. Definition 1.5.3. A non-empty map is called a (p, q)-map if each interior vertex has degree at least p and each interior region has degree at least q.
Small cancellation conditions
Let R be a subset of a free group F . We call R symmetrized if each r ∈ R is cyclically reduced and all cyclically reduced conjugates of r and r−1 are in R. Definition 1.5.4. A word w is called a piece (relative to a symmetrized set R) if it is identically equal to initial segments of two distinct elements in R. Let R be a symmetrized set, λ some positive real number and p a natural number. The following are the hypothesis of small cancellation. 1. C0(λ) : If r ∈ R, r = ws where w is a piece, then `(w) < λ`(r).
2. C(p) : No element of r is a product of fewer than p pieces. In particular we observe that C0(λ) implies C(p) for λ ≤ 1/(p − 1).
3. T (p) : Let 3 ≤ µ < p. Suppose r1, . . . , rµ are the elements of R with no
successive elements riri+1 = 1. Then at least one in r1r2, r2r3, . . . rµ−1rµ, rµr1 is reduced without cancellation.
12 Chapter 1: Preliminaries
Area and Curvature Theorems P Here we mention two theorems which we shall use later. For some notations, M P• will denote summation over vertices or regions in the map M. M will denote summation over boundary vertices or boundary regions in the map M.
Theorem 1.5.5 (Curvature). [70] Let M be a simply-connected (q, p)-map which contains more than one region. Then
• X p + 2 − i(D) ≥ p. (1.8) q M
Theorem 1.5.6 (Area). [70] Let M be a simply-connected [p, q]-map. Then
" • #2 q X V ≤ p − d(v) , (1.9) M p2 M where VM is the total number of vertices in M.
1.5.2 Pictures
Pictures are one of the most powerful tools available in combinatorial group the- ory. Essentially, pictures are the duals of van Kampen diagrams [66]. We will describe pictures briefly as it relates to groups with presentations of the form
G = hX1,X2 | R1,R2,Ri, where G1 = hX1 | R1i, G1 = hX2 | R2i, and R is a word with free product length at least two.
Groups of the form G above are called one-relator products of groups G1 and G2. In the next section we shall discuss such groups in more details. Pictures were first introduced by Rourke [84] and adapted to work for such groups (as G) by Short [89]. Since then they have been used extensively and successfully by various authors in a variety of different ways (see [[18], [19], [21], [34], [56], [57], [63]]). We describe below the basic idea, following closely the account in [62]. A more detailed description can be found in [55] and also [[14], [6], [64], [27], [82]]. Let G be as above, a picture Γ over G on an oriented surface S (usually D2) consists of the following:
1. A collection of disjoint closed discs in the interior of S called vertices;
2. A finite number of disjoint arcs, each of which is either:
(a) a simple closed curve in the interior of S that meets no vertex, (b) an arc joining two vertices (or one vertex to itself),
13 Chapter 1: Preliminaries
(c) an arc joining a vertex to the boundary ∂S of S, or (d) an arc joining ∂S to ∂S;
3. A collection of labels, one at each corner of each region of S (i.e. connected component of the complement in S of the arcs and vertices) at a vertex, and one along each component of the intersection of the region with ∂S. The label
at each corner is an element of G1 or G2. Reading the labels round a vertex in the clockwise direction yields Rn (up to cyclic permutation), as a cyclically
reduced word in G1 ∗ G2.
A region is a boundary region if it meets ∂S, and an interior region otherwise. If S ∼= S2 or if S ∼= D2 and no arcs of meet D2, then Γ is called spherical. In the latter case ∂D2 is one of the boundary components of a non-simply connected region (provided, of course, that Γ contains at least one vertex or arc), which is called the exceptional region. All other regions are interior. The labels of any region 4 of
Γ are required all to belong to either G1 or G2. Hence we can refer to regions as
G1-regions and G2-regions accordingly. Similarly each arc is required to separate a
G1-region from a G2-region. Observe that this is compatible with the alignment of regions around a vertex, where the labels spell a cyclically reduced word, so must come alternately from G1 and G2. A region bounded by arcs that are closed curves will have no labels; nevertheless the above convention requires that it be designated a G1- or G2-region.
2 2 An important rule for pictures on D or S is that the labels within any interior G1- region (respectively G2-region) represents the identity element in G1 (respectively
G2). The label around any given boundary component of the region are formed into a single word read anti-clockwise. Two distinct vertices of a picture are said to cancel along an arc e if they are joined by e and if their labels, read from the endpoints of e, are mutually inverse words in
G1 ∗G2. Such vertices can be removed from a picture via a sequence of bridge moves (see Figure 1.1 and [21] for more details), followed by deletion of a dipole without changing the boundary label.
Figure 1.1: Diagram showing bridge-move.
14 Chapter 1: Preliminaries
A dipole is a connected spherical picture containing precisely two vertices, does not meet ∂S, and none of its interior regions contain other components of Γ. This gives an alternative picture with the same boundary label and two fewer vertices. Let Γ be a picture over G on some surface S. We say that Γ is reduced if it cannot be altered by bridge moves to a picture with a pair of cancelling vertices. If W is a set of words, then Γ is W-minimal if it is non-empty and has the minimum number of vertices amongst all pictures over G with boundary label in W. Any cyclically reduced word in G1 ∗ G2 representing the identity element of G occurs as the boundary label of some reduced picture on D2. A picture is connected if the union of its vertices and arcs is connected. In particular, no arc of a connected picture is a closed arc or joins two points of ∂S, unless the picture consists only of that arc.
Dehn function and Isoperimetric Inequality
In differential geometry, isoperimetric functions are classical. However its use in group theory was originally due to the work of Gromov. In his seminal article [41], Gromov characterised (word) hyperbolic groups as groups with linear isoperimetric functions. We describe these functions and explain the relationship with the solution of the word problem. Throughout this section, we fix a finite presentation
℘ = hX | Ri (1.10) for G. A word w ∈ F (X) represents the identity element in G if and only if it can be expressed as a product of the form
n Y ±1 −1 w = uiri ui , (1.11) i=1 where ui ∈ F (X) and ri ∈ R. We define the area of such w, Area(w), to be the smallest n ≥ 0 such that w can be expressed as a product of the form 1.11.
Definition 1.5.7. A function f : N → N defined by
f(k) = max{Area(w) | w ∈ F (X), w = 1 in G with `(w) ≤ k}, is called the Dehn function or isoperimetric function for the presentation ℘.
Lemma 1.5.8. Let ℘1 and ℘2 be two finite presentations for a group G with corre- sponding Dehn functions f1 and f2. Then for each k ∈ N, there exist constants A,
15 Chapter 1: Preliminaries
B, C, D ∈ N such that
f1(k) ≤ Af2(Bk + C) + D. (1.12)
In particular, if f2 is bounded above by a function that is linear (or quadratic, or polynomial, or exponential, . . . ) in k, then the same is true for f1. These properties are thus invariants of the group G.
Definition 1.5.9. A finitely presented group G has a linear (or quadratic, or poly- nomial, or exponential, . . . ) isoperimetric inequality if for some (and hence for any) finite presentation with Dehn function f, there is a linear (or quadratic, or polynomial, or exponential, . . . ) function g such that f(k) ≤ g(k) for each k ∈ N.
As mentioned earlier, finitely presented groups with linear Dehn functions are called (word) hyperbolic groups. The following theorem shows the connection with solution of the word problem.
Theorem 1.5.10. [31] For a finite presentation ℘ of a group G, the following state- ments are equivalent.
1. G has a recursive isoperimetric function.
2. The word problem is soluble in G.
1.5.3 Combinatorial curvature
For any compact orientable surface (with or without boundary) S with a triangu- lation, we assign real numbers β to the corners of the faces in S. We will think of these numbers as interior angles. A vertex which is on ∂S is called a boundary vertex, and otherwise interior. The curvature of an interior vertex v in S is defined as
" # X κ(v) = 2 − β(v)i π, (1.13) i where the β(v)i range over the angles at v. If v is a boundary vertex then we define " # X κ(v) = 1 − β(v)i π. (1.14) i The curvature of a face ∆ is defined as
" # X κ(∆) = 2 − d(∆) + β(∆)i π (1.15) i
16 Chapter 1: Preliminaries
where β(∆)i are the interior angles of ∆. The combinatorial version of the Gauss- Bonnet Theorem states that the total curvature is the multiple of Euler characteristic of the surface by 2π:
" # X X κ(S) = κ(v) + κ(∆) π = 2πχ(S). (1.16) v ∆ We use curvature to prove results by showing that this value cannot be realised. We assign to each corner of a region of degree k an angle (k − 2)/k. This will mean that regions are flat in the sense that they have zero curvature (alternatively we can make vertices flat instead). This will be the standard assignment for this work. In other words, wherever curvature is mentioned with no specified assignments, it is implicitly assumed that we are using the one described above. In some cases, it may be needful to redistribute curvature (see [23]). This involves locating positively-curved vertices (or regions), and using its excess curvature to compensate its negatively curved neighbours. Hence the total curvature is preserved. We shall describe how to do this in Chapter 6.
1.6 Bass-Serre Theory
Suppose a group G acts on a set X. One can ask if it is possible to reconstruct G (and possibly X and the action) in a way that shows a decomposition theorem for G. For symmetric groups this turns out to be a natural question. It turns out that there is a very good answer when X or its quotient by the action is a tree. In this section we recall some of the ideas about the fundamental theory of Bass and Serre on groups acting freely and without inversion on trees. More details can be found in [[88], [3]]. We begin with the concept of group actions.
1.6.1 Group actions
Definition 1.6.1. Let G be a group and X a non-empty set. A left action of G on X is a map f : G × X → X such that the following holds for each x ∈ X and g, h ∈ G. 1. f(g, f(h, x)) = f(gh, x).
2. f(1, x) = x. In such case, we call X a G-set or a G-space (if we are interested in other properties of X). For simplicity, we shall henceforth write gx instead of f(g, x). For each x ∈ X, we associate to it two sets Gx ⊆ G and Gx ⊆ X.
17 Chapter 1: Preliminaries
Definition 1.6.2. Let X be a G-set. The stabilizer of x ∈ X is the subset
Gx = {g ∈ G | gx = x}.
Definition 1.6.3. Let X be a G-set. The orbit of x ∈ X is the subset
Gx = {y ∈ X | gx = y, g ∈ G }.
Definition 1.6.4. Let X be a G-set. The quotient space (of G action on X) is the set G/X = {Gx | x ∈ X}.
Definition 1.6.5. If X is a G-set and Gx = 1 for all x ∈ G, then we say the action is free.
−1 Lemma 1.6.6. If y = gx, then Gx = (g Gg)y
1.6.2 Group actions on graphs
We recall some notion of graphs.
Definition 1.6.7. A graph X is a pair of sets, V = V (X) 6= ∅ and E = E(X), termed the vertices and edges of X, equipped with three maps
o : E → V, t : E → V, − : E → E, satisfying the following conditions: if e ∈ E, then
1. e 6=e ¯ and e¯ = e;
2. o(e) = t(¯e).
We call o(e), t(e) ande ¯ the origin, terminus and inverse of e respectively. If o(e) = t(e), we call e a loop. Together, we call o(e) and t(e) the extremities of e. Two vertices are adjacent if they are the extremities of some edge. In the plane, graphs are represented by diagrams, with points as vertices and line segments joining its extremities as edges. Usually, an arrow is affixed on such a line segment, whose direction begins from the origin of the edge and directed towards the terminus. When we use diagrams, it is customary to omite ¯.
Definition 1.6.8. Let X1 and X2 be graphs with vertex sets V (X1) and V (X2) respectively. A map φ : V (X1) → V (X2) is called a morphism if it is edge-preserving.
In other words, two vertices in V (X2) are adjacent if their pre-images in V (X1) are adjacent. A morphism is is called an isomorphism if it is a bijection. An isomorphism between V (X1) and itself is called an automorphism
18 Chapter 1: Preliminaries
Definition 1.6.9. A group G acts on a graph X without inversion if g(e) 6=e ¯ for each g ∈ G.
Definition 1.6.10. An orientation of a graph X is a decomposition E(X) = E+ ∪ ¯ ¯ E− of E(X) into two disjoint sets E+ and E− such that E+ = E− and E− = E+.
Since − has order two and is fixed point free, every graph has such an orientation. A graph with a prescribed orientation is called an oriented graph.
1.6.3 Theory of covering spaces
We present briefly some notions in covering space theory, taking for granted many basic notions such as topological space, arc-wise connectedness and the fundamental group π1(X, ∗) of a space X bases at a point ∗. These concepts can be found in most algebraic topology texts such as Massey [74] and Hatcher [45]. For the rest of this section we assume all spaces are Hausdorff i.e., distinct points have disjoint neighbourhoods, and locally arc-wise connected, i.e., if V is an open set containing a point x, there exists an open subset U contained in V and containing x such that any pair of points in U have a path in U joining them.
Definition 1.6.11. Let X and X˜ be two arc-wise connected, locally arc-wise con- nected spaces, p : X˜ → X a continuous map. We call (X,˜ p) a covering space of X if
1. p is onto;
2. each x ∈ X has an open neighbourhood U such that p−1(U) is a disjoint union of open sets homeomorphic via p to U.
Definition 1.6.12. Let X˜ be two arc-wise connected, locally arc-wise connected space. Then a group G acts properly discontinuously on X˜ if there is an action of G on X˜ such that every pointx ˜ ∈ X˜ is contained in an open neighbourhood V satisfying V ∩ gV = ∅ (g ∈ G, g 6= 1.
It is known that if a group G acts properly discontinuously on a suitable space, then G can be recaptured from the quotient space of the action. This idea is the motivation for what we do for the rest of this chapter.
1.6.4 Graph of groups
Suppose a group G acts without inversion on a tree X. We form the quotient graph G/X, keeping track of the stabilisers of some of the vertices and edges in X under the action of G. This information is captured in the so-called graph of groups.
19 Chapter 1: Preliminaries
Definition 1.6.13. A graph of groups (G,X) consists of the following:
1. a connected graph X;
2. a mapping G from V (X) ∪ E(X) into the class of all groups;
3. the image of v ∈ V (X) under G is denoted by Gv and is called the vertex group at v;
4. the image of e ∈ E(X) under G is denoted by Ge and is called the edge group at e;
5. Ge = Ge¯ for every e ∈ E(X);
6. each edge group Ge comes equipped with a monomorphism
Ge → Gt(e) denoted by a 7→ ae.
Given a graph of groups (G,Y ) we associate to it a group which is analogous to the fundamental group of a topological space, called its fundamental group.
1.6.5 The fundamental group of a graph of groups
Let (G,X) be a graph of groups, and T a maximal subtree of X. The following is a definition of the fundamental group π1(G,X) of (G,X).
Definition 1.6.14. [3] Suppose (G,X) and T are as above. Suppose that e ∈ E(T ).
Then we have two monomorphisms Ge → Gt(e) and Ge → Go(e) sending a to ae and a to ae¯ respectively. Define GT to be the group generated by Gv, v ∈ V (T ), with the two images of the edge group Ge in Go(e) and Gt(e) identified according to the prescription ae = ae¯, a ∈ Ge, where e ranges over the edges in T . In other words
GT is simply obtained from the vertex groups, by repeatedly forming amalgamated free products, where the graph of groups (G,X) determines the subgroups to be amalgamated. The group GT can be described in more precise terms as follows.
Choose a vertex v0 ∈ V (T ) and define
L0(T ) = {v0}.
We call L0(T ) the set of vertices at level 0. We then define Ln(T ) to consist of those vertices of T which are at a geodesic distance n from v0. We note that since T is a tree, none of the vertices in Ln(T ) are the extremities of an edge in T . Also since T is connected, ∞ [ V (T ) = Ln(T ). n=0
20 Chapter 1: Preliminaries
Define
G(0) = Gv0 .
We then define G(n+1) inductively, assuming that G(n) has already been defined in such a way that it is generated by all the vertex groups occurring at levels at most n.
For each v ∈ Ln+1(T ) there is a unique edge e ∈ E(T ) such that o(e) ∈ Ln(T ) with t(e) = v. We define G(v) to be the generalised free product of G(n) and Gv with
Ge amalgamated according to the monomorphisms of Ge into the vertex groups at its extremities, as dictated by the graph of groups (G,X). G(n + 1) is then defined to be the generalised free product of all the groups G(v) with G(n) amalgamated. Finally we define ∞ [ GT = G(n). n=0
It follows that GT contains an isomorphic copy of each of the vertex groups Gv (v ∈
V (T ) and that the subgroup of GT generated by Go(e) and Gt(e) is the free product of
Go(e) and Gt(e) amalgamated over Ge for every edge e ∈ E(T ). By inspection on the obvious presentation of GT it is clear that GT does not depend on the choice of v0.
The fundamental group π1(G,X) (with respect to T ) of (G,X), written π1(G,X,T ) is defined to be the HNN extension with possibly infinitely many stable letters. The base group is GT . The choice of the stable latters depends on an orientation, say
E(X) = E+ ∪ E−
of X. Given such an orientation, for each edge e ∈ E+, e 6∈ E(T ) we choose a stable latter te and define
−1 π1(G,X,T ) = hGT , {te | e ∈ E+ − E(T )} | teaete = ae¯ (a ∈ Ge, e ∈ E+ − E(T ))i.
The definition of π1(G,X) given above seems to depend on the choice of T . However it can be shown that this is not the case.
1.6.6 Group action on trees
In this section we state the fundamental theorem of Bass and Serre. We also give examples of groups that act on trees.
Theorem 1.6.15. (Bass-Serre) Let G be a group acting without inversions on a connected tree T . Denote by X the quotient graph G/T . Then there is a graph of groups (G,X)) such that for any maximal subtree T1 of X, G is isomorphic to
π1(G,X,T1). Conversely if (G,X) is a graph of groups, then π1(G,X,T1) acts on a tree in such a way that the resulting graph of groups is isomorphic to (G,X), for
21 Chapter 1: Preliminaries
some choice of maximal tree T1.
We now mention examples of some groups that act on trees. Note that every group acts freely and transitively on its Cayley graph by left multiplication. However, the Cayley graph is not a tree in general.
Example 1.6.16. Let G a free group on two generators x and y. Then the Cayley graph X of G is a tree. The vertex and edge stabilizers of G action on X are trivial. Hence the quotient graph is a graph with one vertex and two edges as shown in Figure 1.2.
Figure 1.2: A graph with two edges and one vertex.
Theorem 1.6.17. A group is free if and only if there exists a tree on which it acts freely.
Example 1.6.18. Let G be the free product G = A ∗ B. Then G acts by left translation on a tree T with E(T ) = G and
V (T ) = {gA | g ∈ G} t {gB | g ∈ G}.
Each edge g = (gA, gB). Clearly the action is free on the edges, but not on the vertices. In particular there is exactly one edge orbit and two vertex orbits GA = −1 {gA | g ∈ G} and GB = {gB | g ∈ G} respectively. Also GgA = gAg ' A, −1 GgB = gBg ' B, and as mentioned before, Gg = 1. Hence the quotient graph is the tree with one edge and two vertices as shown in Figure 1.3.
Figure 1.3: Graph of groups for free product of A and B.
Example 1.6.19. Let G be the free product with amalgamation G = A∗C B. Then G acts by left translation on a tree T with
V (T ) = {gA | g ∈ G} t {gB | g ∈ G}.
22 Chapter 1: Preliminaries
There is an edge e = (gA, hB) between gA and gB whenever hB = gaB for some a ∈ A (alternatively gA = hbA for some b ∈ B). In such case we use gC instead of e. There are two vertex orbits; GA and GB, and one edge orbit. Also GgA = −1 −1 −1 −1 gAg ' A, GgB = gBg ' B, and for gC = (gA, gaB), GgC = gaCa g ' C. Hence the quotient graph is the tree with one edge and two vertices as shown in Figure 1.4.
Figure 1.4: Graph of groups for free product of A and B amalgamated over C.
Example 1.6.20. Let G be the HNN extension of group H with stable letter t and amalgamated subgroup C. Then G acts by left translation on a tree T with V (T ) = {gH | g ∈ G} and E(T ) = {gC | g ∈ G}, where gC = (gH, gtH). All the −1 vertices form a single orbit as well as all the edges. Also GgH = tHt ' H and −1 GgC = gCg ' C. Hence the quotient graph is the tree with one edge and two vertices as shown in Figure 1.5.
Figure 1.5: Graph of groups for HNN extension.
23 Chapter 2
Short review of one-relator product of groups
2.1 Preamble
One can think of Dehn’s most influential contribution to combinatorial group the- ory to be his series of papers in which he posed the algorithmic problems. These problems arose naturally in his study of surface groups or the fundamental groups of two dimensional manifolds. These groups happen to fall into a larger class of groups which are presented with a single defining relator, otherwise known as one-relator groups. The account on one-relator products given here is mostly based on [1]. In this chapter we study these class of groups, their generalisations, known as one- relator products, and some known results about them. We begin with one-relator groups.
2.2 One relator groups
Definition 2.2.1. A one-relator group is a group G whose presentation contains a single defining relator. In other words, G has the form
G = hX | wi.
Some 20 years after Dehn proposed the algorithmic problems, Magnus [71] proved his famous Freiheitssatz.
Theorem 2.2.2. (Freiheitssatz). Let G = hX | wi where w is cyclically reduced. If L is a subset of X which omits a generator occurring in w, then the subgroup of G generated by L is free.
24 Chapter 2: Short review of one-relator product of groups
The Freiheitssatz was so popular that many versions of its proof were produced. These include Lyndon [75] and Schupp[87]. In particular, it led to the first major break-through on the word problem. Theorem 2.2.3. [72] The word problem for one-relator groups has a positive solu- tion. Theorem 2.2.4. [75] Let G = hX | wi where w is cyclically reduced. The group G is torsion free if w is not a proper power. If w = Rn, n > 1, where R itself is not a proper power, then R has order n in G and all elements of finite order of G are conjugates of powers of R. Next we mention another fundamental result known as the Spelling Theorem. Theorem 2.2.5. Let G = hX | Rni where R is cyclically reduced and n > 1. Any non-empty word which represents the identity element must contain a (cyclic) subword of Rn or R−n longer than Rn−1. The Spelling Theorem was first due to Newman [79], and subsequently extended by Gurevich [43], Schupp [87] and Pride [81] (see also [61]).
2.3 One relator product of groups
For an indexing set Λ, let GΛ be the free product ∗γ∈ΛGγ of groups Gγ. Let G be the quotient of GΛ by the normal closure of a word w ∈ GΛ, assumed to be cyclically reduced with length at least two. We write
G = GΛ/N(w). (2.1)
Definition 2.3.1. A group of the form 2.1 is called a one-relator product
Each Gγ is called a factor of G and w is the relator. We say w involves the factor n Gγ if it contains a non-trivial element of Gγ. If w = R for some cyclically reduced word R and n > 1, then we call w a proper power, just like in the case of one-relator groups, with root R. In the case where each factor is isomorphic to the infinite cyclic group, we get a one- relator group. So the question of whether we can prove analogues of results about one-relator groups seems natural. We shall be mostly interested in generalising the Freiheitssatz. For a complete discussion beyond the Freiheitssatz, see the survey article [54] by Howie. We will also mention other results along the line. Definition 2.3.2. Let G be as in 2.1. We say that G satisfies the Freiheitssatz if each Gγ embeds in G by the natural maps. We call Gγ a Freiheitssatz factor if it embeds in G.
25 Chapter 2: Short review of one-relator product of groups
However, things are not so nice in this setting as there is no hope to prove Frei- heitssatz in general. For example, consider G = (A ∗ B)/N(ab), where a and b are elements in A and B respectively with co-prime orders. In this case Freiheitssatz fails since a = b = 1 in G. So to get things to work, more conditions must be imposed. These conditions can be on the factors, the relator or both. We begin with the case of putting condition on the factors.
Theorem 2.3.3. Let G be as in 2.1 with each Gγ locally indicable. Then G satisfies the Freiheitssatz.
Recall that a group is said to be locally indicable if every non-trivial finitely-generated subgroup surjects onto the infinite cyclic group. Theorem 2.3.3 was originally due to Brodskii [[8],[9]]. Later on Howie [51] gave an independent proof. Short [89] gave another proof using pictures. An entirely algebraic proof was given by Baumslag [2]. We mention a few other results about one-relator product of locally indicable groups.
Theorem 2.3.4. [51] Let G = (A ∗ B)/N(w), where A and B are locally indicable groups and w = rm for integer m ≥ 1. No proper subword of w is trivial in G.
Theorem 2.3.5. [61] Let G = (A ∗ B)/N(w), where A and B are locally indicable groups and w = rm for integer m ≥ 1. For every non-empty word s ∈ N(w), either:
1. s is conjugate to w±1, or
2. s contains two almost-disjoint cyclic subwords, each of which is a cyclic sub- word of w±1 longer than rm−1.
As a corollary to Theorem 2.3.5, we have that the solubility of the word problems for A and B implies that of G. It seemed at the time that torsion-freeness was the nice property we require of the factor groups to make things work. In particular we have the following conjecture.
Conjecture 2.3.6. The Freiheitssatz holds for one-relator products with torsion-free factors.
This conjecture is still widely open. However there are some results supporting it. We mention one next and another when we discuss equation over groups.
Theorem 2.3.7. [10] Let G = (A ∗ B)/N(w) where A and B are torsion-free groups and w = a1b1 . . . akbk is a cyclically reduced word in A ∗ B such that some ai is isolated in (a1, . . . , ak) and some bj is isolated in (b1, . . . , bk). Then G satisfies the Freiheitssatz.
26 Chapter 2: Short review of one-relator product of groups
We say that xi is isolated in (x1, . . . , xk) if xj 6∈ hxii for j 6= i. Next consider the case where a condition is put on the relator instead. Predominant among these is that the relator w is a proper power i.e. G = (A ∗ B)/N(w) where w = Rn for some n > 1. It turns out that the higher n is, the easier it is to handle. For instance if n > 6, then Freiheitssatz for G follows from small-cancellation results of Lyndon [39] (see also [[70],Chapter V]), since the symmetrized closure of Rn satisfies the metric small-cancellation condition C0(1/6). More generally the following result holds.
Theorem 2.3.8. Let G = (A ∗ B)/N(w), where w = Rn. Then G satisfies the Freiheitssatz if n ≥ 4, or n ≥ 3 and R has no element of order 2.
The case where n = 6 is due to Gonzalez-Acuna and Short [34], using pictures. It is also a consequence of [[15] Theorem 5.1] due to Collins and Perraud. The cases of n = 4, 5 are due to Howie [[55], [56]], again by use of pictures. The remaining case is due to Howie and Duncan [20].
Conjecture 2.3.9. [53] Let G = (A ∗ B)/N(w), where w = rn and n > 1. Then G satisfies the Freiheitssatz.
Most of the results proved above are either geometric or combinatorial. We move on to a special class of one-relator products where results are proved using repre- sentation theory.
2.4 One relator product of cyclics
In the next chapter we will be studying one-relator product induced from generalised triangle groups. Hence it is important to recall what generalized triangle groups are. But first we describe the larger family of one-relator products of cyclics. These are groups of the form
e1 e2 en G = hx1, x2, . . . , xn | x1 = x2 = ... = xn = w(x1, x2, . . . , xn) = 1i, (2.2) where w = w(x1, x2, . . . , xn) is cyclically reduced and ei > 1 for each i = 1, 2, . . . , n. In other words G is a one-relator product in which each factor is finite cyclic. We shall only be interested in the case where w is a proper power - say w = Rm with m > 1, so henceforth we assume that we are in this situation. One-relator product of cyclics provides a natural generalization of discrete subgroups of PSL2(R), otherwise known as Fuchsian groups. A very important special case is when n = 2, i.e.
27 Chapter 2: Short review of one-relator product of groups
e1 e2 m G = hx1, x2 | x1 = x1 = R(x1, x2) = 1i (2.3)
A group of the form 2.3 is called a generalised triangle group. When R(x1, x2) = x1x2 (up to permutation and inversion) we get the ordinary triangle group. These groups have been much studied and a lot is known about them. For exam- ple Baumslag, Morgan and Shalen used a technique based on the idea of Ree and Mendelsohn [83] to prove the following. Theorem 2.4.1. [4] A generalised triangle group admits an essential representation into PSL2(C). Recall that the generalised triangle group in 2.3 is said to admit an essential rep- resentation ρ : G → H if in H, ρ(xi) has order ei and ρ(R) has order m. So in particular, this shows that not only is the group non-trivial, the factor groups all embed. Theorem 2.4.1 was independently proved by Boyer [7] who produced an essential representation in SU(2). A variant of Boyer’s method was utilized by Howie [58] to prove the non-triviality of one-relator product with three cyclic fac- tors. However much more was possible using representations in PSL2(C). Indeed many problems about generalised triangle groups tackled using this technique. One of such problems is the Rosenberger conjecture which states that generalised trian- gle groups either contains a non-abelian free group or a solvable subgroup of finite index. In other words, generalised triangle groups satisfy the Tits Alternative. So it makes sense to recap the basic idea.
Suppose X1 and X2 are matrices in SL2(C) with traces Tr(X1) and Tr(X2) respec- tively, and W = W (X1,X2) is a word in X1 and X2. Then the trace Tr(W ) of W is a polynomial in the variables Tr(X1), Tr(X2) and Tr(X1X2) (the trace of X1X2) [49]. Using this an essential representation of the generalised triangle group G in 2.3 can be constructed.
It is known that an element X ∈ PSL2(C) has order e > 1 if and only if Tr(X) =
2 cos(kπ/e) for some integer k co-prime to e. Hence the images of x1 and x2 can be forced to have orders e1 and e2 by mapping them to matrices X1 and X2 with traces
2 cos(π/e1) and 2 cos(π/e2) respectively. Then the trace Tr(W ) of W (the image of m R (x1, x2)) becomes a one-variable polynomial in Tr(X1X2) of degree m`(R(x1, x2)).
An essential representation of G is obtained by choosing Tr(X1X2) to be the root of Tr(W ). A refinement of the technique of Baumslag, Morgan and Shalen shows that it is sufficient for the factor groups to have faithful representation in PSL2(C).
Theorem 2.4.2. [28] Let A and B be groups with faithful representations in PSL2(C), and suppose R ∈ A∗B is cyclically reduced with `(R) ≥ 2. Then G = (A∗B)/N(Rm),
28 Chapter 2: Short review of one-relator product of groups
where m > 1, has a representation ρ : G → PSL2(C) such that ρ is faithful on each factor and ρ(R) has order m. In particular the Freiheitssatz holds for G
In chapter 4 we shall be proving various results similar to Theorem 2.4.2. Another result about generalised triangle groups which will be very useful in the analysis of the so-called clique-pictures is the following.
Theorem 2.4.3. [60](Spelling Theorem for generalised triangle groups) For a gen-
p q r Qk αi βi eralised triangle group G = h x, y | x , y ,W (x, y) i, where W (x, y) = i=1 x y , Ql γi δi (k > 0, 0 < αi < p, 0 < βi < q). If V (x, y) = i=1 x y , (l > 0, 0 < γi < p, 0 <
δi < q) is a trivial word in G, then l ≥ kr.
Suppose we have two subgroups G1 and G2 of G, with the natural inclusion maps
G1 → G and G2 → G. We get a unique map f : G1 ∗G2 → G. The Gersten-Stallings angle (G; G1,G2) [91] is defined to be:
1. 0, if f is injective, or
2. π/n, where 2n is the minimal length of a non-trivial element in the kernel of f.
In general, it is not easy to calculate these angles. However, the Spelling Theorem gives a lower bound for generalised triangle groups. Finally before we end this section, we mention the following result of Fine, Howie and Rosenberger [28] about one-relator product of cyclics.
Theorem 2.4.4. Let G be as in 2.2. Then G admits an essential representation
ρ : G → PSL2(C) such that ρ restricted to any subgroup generated by any proper subset of {x1, x2, . . . , xn} is faithful. In particular Freiheitssatz holds for G and any proper subset of {x1, x2, . . . , xn} generates the obvious free product.
2.5 Relative presentations
Let H be a an arbitrary group, and F a free group. A relative presentation is a one-relator product of the form
(G ∗ F ) H = , (2.4) N(w) where w is a word in G ∗ F . We will only discuss the case where F is a free group of rank 1 with generator t.
Conjecture 2.5.1. (Kervaire) H is non-trivial if G is non-trivial.
29 Chapter 2: Short review of one-relator product of groups
This conjecture has been proved for a large class of groups, for example when G is locally residually finite [32] or locally indicable [[8], [50], [89]]. It is still open in general (even in the case where F is an arbitrary group) whether or not H is trivial. We think of w = 1 as an equation with coefficients in G and variable t. We say that the equation w = 1 has a solution in G if the identity map on G extends to a homomorphism from H ∗ hti to G with w in its kernel. We say that w = 1 has a solution over H if H can be embedded in a group G˜ such that w = 1 has a solution in G˜. This property is equivalent to the canonical map from G to H being injective.
Problem 2.5.2. (Adjunction) Under what conditions does w = 1 have a solution over G.
In general it is not possible to find a solution to an arbitrary equation w = 1 over an arbitrary group G. As in the case of Freiheitssatz, some sort of restriction has to be placed on the group, on the equation, or both. For example the equation xtyt−1 = 1 has no solution in any group if x and y have co-prime orders. Also there is no chance if w is an element of G. So a possible restriction on the equation is that the w is non-singular, in other words the exponent sum of w in t is non-zero. Even with this restriction on the w, it is still a widely open problem whether a solution over G always exists. There is a positive answer when w is non-singular and G is locally residually finite or locally indicable. Klyachko gave a positive answer in the case where G is torsion-free and w has exponent sum ±1 in t, thereby providing further evidence in favour of Conjecture 2.3.6. Levin [67] showed that a solution exists if t occurs in w only with positive exponents. Some other works show that the result holds when the sum of exponent sums of t in w is small [see [22], [52], [26]].
30 Chapter 3
One-relator product induced from generalised triangle group
3.1 Preamble
In this chapter we consider one-relator product of two non-trivial groups G1 and G2 of the form
(G ∗ G ) G = 1 2 , (3.1) N(Rn) where R is contained in a subgroup A∗B of G1 ∗G2 with A and B conjugate to cyclic subgroups of G1 and/or G2, and natural number n ≥ 2. We can of course assume that A is generated by some element a and B is generated by UbU −1 for some word −1 U and letter b in G1 ∗ G2. Hence R is a word in the subgroup hai ∗ hUbU i of
G1 ∗ G2 which we can assume to be cyclically reduced. As in [62], we also require in some cases the technical condition that (a, b) be admissible: whenever both a and b belong to same factor, say G1, then either the subgroup of G1 generated by {a, b} is cyclic or hai ∩ hbi = 1. If a one-relator product is in the form of G described above, then we say G is induced by a generalised triangle group. In such a case G can be realised as a push-out of groups as shown in Figure 3.1,
A ∗ B H
G1 ∗ G2 G
Figure 3.1: Push-out diagram.
31 Chapter 3: One-relator product induced from generalised triangle group where H is the corresponding generalised triangle group
H = hx, y | xp, yq,R0(x, y)n i, such that x 7→ a, y 7→ UbU −1 and R0(x, y) 7→ R. One-relator product induced by a generalised triangle group crops up in different forms when proving results. One frequent place is in the analysis of zones (to be defined later) of size `(W )/2 or `(W ), connecting vertices of same sign, when proving results about groups with one-relator presentation (with relator W ) using pictures. Some of the references where such groups have cropped up include [[21], [44], [55], [56], and [57]]. However, the group only earned its name in [62], where a number of results were proved under the hypotheses that n ≥ 3 and that the pair (a, b) is admissible. Here we prove similar results under hypotheses that are in general weaker than those assumed in [62]. Hypothesis A. n ≥ 2, R has length at least 4 as a word in the free product hai ∗ hUbU −1i, and the pair (a, b) is admissible.
Hypothesis B. n ≥ 2 and no letter of R (as a word in G1 ∗ G2) has order 2 . Under either of the above hypotheses, we prove the following.
Theorem 3.1.1. The natural maps G1 → G, G2 → G and H → G are all injective, where H is the quotient of hai ∗ hUbU −1i by N(Rn).
Theorem 3.1.2. If the word problems are soluble for H, G1 and G2, then it is soluble for G.
n Theorem 3.1.3. If some cyclic permutation of R has the form W1W2 with 0 < n `(W1), `(W2) < `(R ) as words in G, then W1 6= 1 6= W2 as words in G. In particular R has order n in G.
The first part of Theorem 3.1.1 is a generalisation of Magnus’ Freiheitssatz for one- relator groups [71]. There are many generalisations of the Freiheitssatz to one-relator products of a special nature. Theorem 3.1.3 is a version of a result of Weinbaum [95] for one-relator groups. All three of these results were proved in [62] under the hypotheses that n ≥ 3 and that the pair (a, b) is admissible. The above theorems will be proven using pictures and clique-picture (see Section
3.4). Throughout this Chapter W denotes the set of non-trivial elements in G1 ∪ −1 G2 ∪ hai ∗ hUbU i.
32 Chapter 3: One-relator product induced from generalised triangle group
3.2 Periodicity
Definition 3.2.1. A word w of length n has a period γ if γ ≤ n and wi = wi+γ for all i ≤ n − γ.
Example 3.2.2. Suppose that w = x1x2x3x1x2 is a group G. Then w has period 3. Definition 3.2.3. A word w is said to be bordered by u and v if u and v are proper initial and terminal segments of w respectively. Furthermore we say that w is bordered by u if u ≡ v. Remark 3.2.4. It follows immediately that a w bordered by u has period γ = `(w) − `(u). Theorem 3.2.5. [29] Let w be a word having periods γ and ρ with ρ ≤ γ. If `(w) ≥ γ + ρ − gcd(γ, ρ), then w has period gcd(γ, ρ). Even though Theorem 3.2.5 was originally due to Fine and Wilf, there has been other proofs. In particular we shall include a proof as given in [[55] Proposition 1]. We find this version useful as many of the arguments presented later not only uses this result, but are constructed in a similar way.
Proof of Theorem 3.2.5. Let γ = sν and ρ = tν where ν = gcd(γ, ρ). If ν ∈ {γ, ρ}, then there is nothing to show. Hence we assume without loss of generality that s > t > 1. Let V denote the initial segment of w of length γ. We express V as
V = V1V2 ...Vs (3.2)
with each Vi of length ν. Hence to show that w has period ν, it suffices to show that
V1 = V2 = ... = Vs. (3.3)
By hypothesis, w has an initial segment of the form
V1V2 ...VsV1V2 ...Vt−1. (3.4)
By period ρ = tν of w, we have that Vi = Vi+t for 1 ≤ i ≤ s − 1 (all indices modulo s). This gives s − 1 equations which involves each Vi for i = 1, 2, . . . , s. For suppose that i ≡ i + mt modulo s for 0 < m < s, then m ≡ 0 modulo s (since s and t are co-prime). But this implies that m = 0, which is impossible.
Corollary 3.2.6. Let w be a word having initial segment w1 with period γ and terminal segment w2 with period ρ. If w1 and w2 intersect in a segment u with `(u) ≥ γ + ρ − gcd(γ, ρ), then w has period gcd(γ, ρ).
33 Chapter 3: One-relator product induced from generalised triangle group
Lemma 3.2.7. Suppose that W is a word in the free monoid S∗ equipped with an involution, s 7→ s−1, possibly with fixed points. Suppose that W is a word in S∗ the form
−1 W = x1V1y1V1 = z0z1 ··· z2k−1, for some letters x1, y1 and some word V1, where `(W ) = 2k. Suppose also that W has a cyclic permutation of the form
−1 zjzj+1 ··· z2k−1z0 ··· zj−1 = x2V2y2V2 , for some letters x2, y2 and some word V2, where j 6≡ 0 modulo k. Then one of the following holds:
1. {x1, y1} = {x2, y2} and
s Y α(j) β(j) −1 W ≡ [x1 V3y1 V3 ], (3.5) j=1
for some odd integer s > 1 and some word V3, with α(j), β(j) = ±1 for each j.
−1 2. yi = xi for i = 1, 2, and
s Y α(j) β(j) −1 W ≡ [x1 V3x2 V3 ], (3.6) j=1
for some even integer s > 0 and some word V3, with α(j), β(j) = ±1 for each j.
Proof. In what follows all subscripts are counted modulo 2k. By hypothesis we −1 −1 have x1V1y1V1 = z0z1 ··· z2k−1 and so zi = z2k−i unless i ≡ 0 modulo k. Similarly −1 zi = z2j−i unless i ≡ j modulo k.
Let V3 = z1z2 ··· zm−1 where m = gcd(j, k). For i 6≡ 0 modulo m, we have zi = −1 z2j−i = zi−2j and also zi = zi+2k, and so zi = zi+2m. It follows that
s −1 Y −1 x1V1y1V1 = [ξtV3ηtV3 ] (3.7) t=1 for some letters ξt, ηt, where s = k/m. By hypothesis j 6≡ 0 modulo k, and so s > 1. Suppose first that s is odd. Replacing j by k + j if necessary, we may assume that
34 Chapter 3: One-relator product induced from generalised triangle group j/m is also odd. This gives a chain of equalities
−1 z0 = z2j = z2k−2j = ..., (3.8)
±1 which continues until it reaches zd for some subscript d ∈ {j, k, j + k}. Since the equalities link letters with subscripts of the same parity, we must have d = j + k.
Moreover, every ze with e ≡ 0 modulo 2m appears in the chain. There are precisely s such letters, so an even number of equalities, and hence ξ1 = x1 = z0 = zj+k = y2 ±1 ±1 ±1 ±1 and ξt = x1 = y2 for each t. By a similar argument ηt = y1 = x2 for each t. Now suppose that s is even. Then j/m is odd. Arguing as above, we have a chain of s − 1 equalities −1 z0 = z2j = ..., (3.9)
−1 −1 which must end with zk , and a similar chain of equalities equating zj with zj+k. ±1 ∓1 ±1 ∓1 Hence in this case ξt = x1 = y1 for each t, and ηt = x2 = y2 for each t.
3.3 Generalised triangle group description and re- finements
Let R be a cyclically reduced word of length at least 2 in the free product G1 ∗ G2. As mentioned in the introduction, we are interested in the case where R is contained in the subgroup A ∗ B, where A and B are cyclic subgroups of conjugates of G1 or n n G2. Let N(R ) denote the normal closure of R in G1 ∗ G2 with n ≥ 2. Then the group of interest is the following:
(G ∗ G ) G = 1 2 N(Rn)
If S and T are the generators of A and B respectively, then we can construct a generalised triangle group H = hx, y | xp, yq,R0(x, y)n i, such that x 7→ S, y 7→ T 0 0 and R (x, y) 7→ R (S,T ) (freely equal to R in G1 ∗ G2). The group G can then be realised the push-out in Figure 3.1. We will call the set {S,T } or Figure 3.1 the choice of generalised triangle group description for G
The subgroup A ∗ B of G1 ∗ G2 may not be unique amongst all two-generator sub- 0 0 groups containing R. If A ∗ B is another two-generator subgroup of G1 ∗ G2 with suitable generating set {S0,T 0} and A ∗ B ⊆ A0 ∗ B0, then we can write S, T and R0(S,T ) as words in this new generating set. In general we have that
`(S0) + `(T 0) ≤ `(S) + `(T ) (3.10)
35 Chapter 3: One-relator product induced from generalised triangle group and `(R0) ≤ `(R00) where R ≡ R0(S,T ) ≡ R00(S0,T 0). If any of the two inequalities is strict, we say that the generalised triangle group description given by R00 is a refinement of the one given by R0. Let p0 and q0 be the orders of S0 and T 0 respectively. Then we have that the group H0 = h x0, y0 | xp0 , yq0 ,R00(x, y)n i and the refinement gives a commutative diagram as in Figure 3.2, in which both squares are push-outs.
Cp0 ∗ Cq0 Cp ∗ Cq G1 ∗ G2
H0 H G
Figure 3.2: Double push-out diagram.
A generalised triangle group description for G is said to be maximal if no refinement is possible. In other words, A ∗ B is maximal (with respect to inclusion) amongst all two-generator subgroups of G1 ∗ G2 containing R. It follows from Equation 3.10 that maximal refinements always exist (but are not necessarily unique). From now on we will assume that we are working with maximal refinements. We next note a useful consequence in this context of Lemma 3.2.7.
Lemma 3.3.1. Suppose that, in the above, the generators S and T of A and B respectively have the forms S = a, T = UbU −1 for some letters a, b and some word U. Suppose that (a, b) is an admissible pair. If there are integers α, β, γ, δ such that α β −1 γ δ −1 a Ub U and a Ub U are proper cyclic conjugates in G1 ∗ G2, then a refinement is possible.
Proof. We may apply Lemma 3.2.7 to aαUbβU −1 and aγUbδU −1 except in the situa- tion where aαUbβU −1 ≡ bδU −1aγU. Let us first consider this exceptional situation. Then in particular U ≡ U −1 and so U has the form V xV −1 for some word V and some letter x of order 2. But we also have aα = bδ, and so by definition of ad- missibility a, b have a common root c say. But then A ∗ B is a proper subgroup of −1 C ∗ D, where C and D are the cyclic subgroups of G1 ∗ G2 generated by c, V xV respectively. This is a refinement, as required.
α β Hence we are in a situation where we can apply Lemma 3.2.7 with x1 = a , y1 = b , γ δ x2 = a , y2 = b , and V1 = V2 = U. There are two possibilities to consider depending on the parity of s.
36 Chapter 3: One-relator product induced from generalised triangle group
If s is odd in the conclusion of Lemma 3.2.7. In this case, A∗B is a proper subgroup 0 0 −1 of A ∗ B , where B is the cyclic subgroup generated by V3bV3 , so again we have a refinement. Otherwise if s is even in Lemma 3.2.7. Then aα = b−β, so by admissibility a, b have a common root c. Then A ∗ B is a proper subgroup of C ∗ D, where C and D are −1 the cyclic subgroups generated by c and V3cV3 respectively. As before this gives a refinement.
3.4 Clique-pictures
Clique-pictures appeared in [62] and are modelled on generalised triangle groups. For the rest of this Thesis, (G ∗ G ) G = 1 2 N(Rn) is a one-relator product induced by the generalised triangle group
H = hx, y | xp, yq,R0(x, y)ni.
−1 In other words R is a word in {a, UbU } for some word U ∈ G1 ∗ G2 and letters a and b in G1 ∪ G2 with orders p and q respectively. If u and v are two vertices in a picture over G that are joined by an arc e, then we may use the endpoints of e as the starting points for reading the labels Lu and Lv of u and v respectively. In each case the label is a cyclic permutation of R0(a, UbU −1)±n. We may assume, without loss of generality, that the word R0(x, y) begins with the letter x. Choose a cyclic permutation R∗(x, y) of R0(x, y)−1 that also starts with x±1.
−1 0 −1 n ∗ −1 n Each of Lu and Lv is a cyclic conjugate of R (a, UbU ) or R (a, UbU ) , say 0 −1 n ∗ −1 n −1 0 0 Lu = YZ, where ZY = R (a, UbU ) or ZY = R (a, UbU ) and Lv = Y Z , where Z0Y 0 = R0(a, UbU −1)n or Z0Y 0 = R∗(a, UbU −1)n. We u and v equivalent written u ∼ v, if and only if `(Y 0) ≡ `(Y ) modulo l = `(aUbU −1). It follows immediately from Lemma 3.3.1 that `(Y 0) and `(Y ) are unique modulo l, and so the relation ∼ is well-defined. The point of the relation ∼ is that, when u ∼ v, then the 2-vertex sub-picture consisting of u and v, joined by e and any arcs parallel (see Remark 3.4.4 for definition) to e, has boundary label a word in {a, UbU −1}, after cyclic reduction and cyclic permutation. Indeed, the cyclic reduction of the label can be achieved by performing bridge moves to make the number of edges parallel to e be a multiple of l/2 (see Figures [3.3, 3.4, 3.5]). Now
37 Chapter 3: One-relator product induced from generalised triangle group let ≈ denote the transitive, reflexive closure of ∼. Then ≈ is an equivalence relation on vertices. After a sequence of bridge moves, we may assume that arcs joining equivalent vertices do so in parallel classes each containing a multiple of l/2 arcs (see Figures [3.6, 3.14]).
As an illustration, suppose U = u1u2 . . . ul/2−1 and u and v are vertices of a picture with labels k Y label(u) = aαi Ubβi U −1 (3.11) i=1 and k Y label(v) = aγj Ubδj U −1 (3.12) j=1 respectively. Suppose also that u and v are as shown in Figure 3.3 below.
Figure 3.3: Diagram showing an example of u ∼ v.
Then by definition u ∼ v. After doing bridge-moves u and v are joined by l/2 parallel class of arcs as shown in Figure 3.4.
Figure 3.4: Diagram showing u and v joined by l/2 arcs after doing bridge-moves to Figure 3.3.
38 Chapter 3: One-relator product induced from generalised triangle group
Note that this l/2 parallel class of arcs is special in the sense that the label on u contained in it is U and that of v is U −1. We replace such class by a shaded rectangle as shown in Figure 3.5.
Figure 3.5: Diagram showing the l/2-zone in Figure 3.4 replaced by a single shaded rectangle.
Similarly if u and v are two equivalent vertices joined by a multiple of l/2 parallel class (possibly after doing bridge-moves) as shown in Figure 3.6, we replace such class by a shaded rectangle as shown in Figure 3.14.
Figure 3.6: Diagram showing two vertices u ∼ v joined by l-zone.
Figure 3.7: Diagram showing the l-zone in Figure 3.6 replaced by a single shaded rectangle.
39 Chapter 3: One-relator product induced from generalised triangle group
Definition 3.4.1. A clique is a sub-picture consisting of any ≈-equivalence class of vertices, together with all arcs between vertices in that ≈-class (assumed to occur in parallel classes of multiples of l/2 arcs), and all regions that are enclosed entirely by such arcs.
In other words vertices that can be joined by shaded rectangles make up the vertices in a single clique.
Definition 3.4.2. A clique is said to be simply-connected if these collection of shaded rectangles in it forms a tree or a simple closed curve (i.e. a curve with no self-intersections).
Figure 3.8: Diagram showing a simply-connected clique consisting of only two ver- tices and how the two vertices are combined to form one single vertex.
The point here is that vertices in a simply-connected clique can be combined along the shaded rectangle (see Figure 3.8) to form a single vertex, which we shall also call a clique. Note that we can have cliques which are not simply-connected. For example a clique containing the vertices v1, v2, . . . , v5 as shown in Figure 3.9 below is not simply-connected.
Figure 3.9: Diagram showing a non-simply connected clique.
40 Chapter 3: One-relator product induced from generalised triangle group
This leads us to the notion of Clique-picture
Definition 3.4.3. Let G be a one-relator product induced from a generalised trian- gle group as above, and let P be a picture on a surface Σ, such that every clique of P is simply-connected. Then the clique-quotient of P is the picture formed from P by contracting each clique to a point, and regarding it as a vertex. A clique-picture P over G is the clique-quotient of some (reduced) picture over G. The label of a vertex in a clique-picture is called a clique-label.
The process of joining two vertices of P or two cliques of P to form a single clique is called amalgamation. (Here we also include the possibility of amalgamating a clique with itself. By this we mean adding arcs from v to v and/or regions to an existing clique v, which could alter some properties of the clique such as simple-connectivity.) If it is possible to amalgamate two cliques in P(possibly after doing bridge-moves), we say that P is not reduced, and reduced otherwise. A picture whose vertices are cliques is said to be W-minimal if it is non-empty and has the minimum number of cliques amongst all pictures over G with boundary label in the set W. Remark 3.4.4. Let Γ be P or P. Two arcs of Γ are said to be parallel if they are the only two arcs in the boundary of some simply-connected region 4 of Γ. We will also use the term parallel to denote the equivalence relation generated by this relation, and refer to any of the corresponding equivalence classes as a class of ω parallel arcs or ω-zone (see Figure 3.10).
Figure 3.10: Diagram showing vertices v1 and v5 joined by a 2-zone and vertices v2 and v5 joined by a 3-zone.
Given a ω-zone joining vertices u and v of Γ, consider the ω − 1 two-sided regions separating these arcs. Each such region has a corner label xu at u and a corner label xv at v, and the picture axioms imply that xuxv = 1 in G1 or G2. The ω − 1 corner labels at v spell a cyclic subword s of length ω − 1 of the label of v. Similarly the corner labels at u spell out a cyclic subword t of length ω − 1. Moreover, s = t−1. If we assume that Γ is reduced, then u and v do not cancel. Hence the
41 Chapter 3: One-relator product induced from generalised triangle group cyclic permutations of the labels at v and u of which s and t are initial segments respectively are not equal. Hence t and s are pieces. As in graphs, the degree of a vertex in Γ is the number of zones incident on it. For a region, the degree is the number corners it has. We say that a vertex v of Γ satisfies the local C(m) condition if it is joined to at least m zones. We say that Γ satisfies C(m) if every interior vertex satisfies local C(m).
3.5 Clique-labels and virtual periodicity
In this section we obtain some preliminary results about clique-pictures. One ad- vantage clique-pictures has over ordinary pictures is that some cyclic permutation of the inverse of any clique-label can also be interpreted as a clique-label. Thus we may regard any clique as having either possible orientation, as convenient. We make the convention that all our cliques have the same (positive or clockwise) orientation. Throughout this Chapter we shall assume that our clique-picture is W-minimal. Note that up to cyclic permutation the clique-label of a clique u has the form
k Y label(u) = aαi Ubβi U −1 (3.13) i=1 for 0 < αi < p and 0 < βi < q. Denote our clique-picture by Γ and let v be a clique of Γ. Take a cyclic permutation c(k) of the label of v of the form (3.13) and express it as
c(k) = z0z1 ··· zkl−1 (3.14) where l = `(aαi Ubβi U −1).
We call a letter zj of a clique-label
k Y αi βi −1 label(u) = a Ub U = z0z1 ··· zkl−1 (3.15) i=1 special if j ≡ 0 modulo l/2. Note that every special letter is equal to a power of a or of b.
Let Ω := (G1 ∪G2)\{1}. Define ∼ to be the smallest equivalence relation on Ω with the property that aα ∼ a for all α such that aα 6= 1 and bβ ∼ b for all β such that bβ 6= 1. Note that the natural involution x 7→ x−1 on Ω descends to an involution on Ω/ ∼, under which the ∼-classes of a and b are fixed points. We will sometimes work in the free monoid (Ω/ ∼)∗. In particular clique-labels are periodic in (Ω/ ∼)∗ with period l.
42 Chapter 3: One-relator product induced from generalised triangle group
Let v be a clique of degree k. This means that there are k zones incident at v, say
Z1,Z2,...,Zk labelled consecutively in clockwise order around v as shown in Figure 3.11.
Figure 3.11: Diagram showing a vertex v of degree five.
Recall that each zone Zi is a class of parallel arcs. The number of arcs in Zi is denoted by ωi. If Zi connects cliques u and v (possibly u = v), then Zi determines cyclic subwords si and ti of length ωi − 1 of the clique-labels of u and v respectively, −1 such that si ≡ ti in G1 ∗ G2
Definition 3.5.1. A zone Zi is said to be large if ωi > l/2 and small otherwise.
We will use the following generalisation of the concept of periodic word, as applied to cyclic subwords of
n Y α(k) β(k) −1 label(v) = z0z1 ··· znl−1 = [a Ub U ]. (3.16) k=1
Definition 3.5.2. We say that a cyclic subword W = zj ··· zk (subscripts modulo nl) of label(v) is virtually periodic with virtual period µ if, for each i ∈ {j, j + 1, . . . , k − µ}, one of the following happens:
1. zi = zi+µ;
ψ 2. a special letter zd = a belongs to W , for some d ≡ 0 modulo l, i ≡ d modulo
µ, and each of zi, zi+µ is equal to a power of a;
ψ 3. a special letter zd = b belongs to W , for some d ≡ l/2 modulo l, i ≡ d modulo
µ, and each of zi, zi+µ is equal to a power of b;
ψ 4. a and b have a common root c in G1 or G2, a special letter zd = c belongs to
W , for some d ≡ 0 modulo l/2, i ≡ d modulo µ, and each of zi, zi+µ is equal to a power of c.
43 Chapter 3: One-relator product induced from generalised triangle group
Recall that the pair (a, b) is assumed to be admissible. If a and b have a common power in G1 and G2, then they have a common root, and in that case the second and third possibilities in the above definition are subsumed in the fourth. Otherwise the fourth possibility cannot occur. By definition, the clique-label label(v) itself is virtually periodic of virtual period l. Other examples of virtually periodic words arise from zones incident at v.
Lemma 3.5.3. Suppose that Zi is a zone incident at v. Then there is a positive + integer µ ≤ l/2 and a cyclic subword si of label(v) of length ωi + µ − 1 and virtual + period µ, such that si is either an initial or a terminal segment of si .
Proof. Let si be the cyclic subword zjzj+1 ··· zk of
n Y α(k) β(k) −1 label(v) = z0z1 ··· znl−1 = [a Ub U ]. (3.17) k=1
−1 The zone Zi links v to an adjacent clique u and identifies si with ti for some −1 −1 cyclic subword ti of label(u). Thus ti is a cyclic subword of label(u) . Write −1 ti = yj0 yj0+1 ··· yk0 where
m −1 Y γ(k) δ(k) −1 label(u) = y0y1 ··· yml−1 = [a Ub U ]. (3.18) k=1
−1 0 0 Since si ≡ ti , then in particular k − j ≡ `(ti) − 1 = `(si) − 1 ≡ k − j modulo l. If j ≡ j0 modulo l, then we may amalgamate the cliques u and v, contrary to hypothesis. Hence there are integers q and µi such that 0 ≤ q ≤ m, 0 < µi ≤ l/2 0 such that j = j + ql ± µi. Define ( z ··· z if j0 = j + ql − µ s+ = j−µi k i i 0 zj ··· zk+µi if j = j + ql + µi
+ In the first case, si is a terminal segment of si , while the initial segment of the same −1 length agrees with ti ≡ si, except possibly at special letters zd (d ≡ 0 modulo l/2) which may be a different power of a (or of b) than the corresponding letter of si. It + follows that si is virtually periodic of virtual period µi, as claimed.
The second case is entirely analogous, except that si is an initial rather than a + terminal segment of si .
We need to analyse the interaction of virtually periodic subwords of label(v) obtained by applying Lemma 3.5.3 to two adjacent large zones at v. To do this we will use the following analogue of Corollary 3.2.6.
44 Chapter 3: One-relator product induced from generalised triangle group
Lemma 3.5.4. Suppose that the cyclic subword W = zj ··· zk (subscripts modulo nl) of n Y α(k) β(k) −1 label(v) = z0z1 ··· znl−1 = [a Ub U ]. (3.19) k=1 is the union of a virtually periodic segment W1 of virtual period µ and a virtually periodic segment W2 of virtual period ν. Let γ = gcd(µ, ν). If the intersection of these segments has length at least µ + ν − γ, then W is virtually periodic of virtual period γ.
Proof. Let i, m be such that zi and zi+mγ are letters of W . Then we claim there is a finite chain of subscripts i(0), i(1), . . . , i(N) with i(0) = i and i(N) = i + mγ such that, for each t either
|i(t) − i(t + 1)| = µ (3.20) and zi(t) and zi(t+1) are letters of W1, or
|i(t) − i(t + 1)| = ν (3.21) and zi(t) and zi(t+1) are letters of W2.
Certainly each letter in W1 (respectively W2) is linked to some letter in W0 :=
W1 ∩ W2 by such a chain, since `(W0) ≥ max(µ, ν), so it suffices to prove the claim when zi, zi+mγ are letters of W0. Write mγ = αµ + βν where α, β ∈ Z, and argue by induction on |α| + |β|. Without loss of generality, assume that α > 0. If zi+µ is a letter of W0, then the result follows by applying the inductive hypothesis to zi+µ, zi+mγ. Otherwise, β < 0 and zi−ν is a letter of W0, so we may apply the inductive hypothesis to zi−ν, zi+mγ. This proves the claim. Now we take m = 1 in the above, and prove that at least one of the alternative conditions for virtual periodicity holds.
If zi(t) = zi(t+1) for all n then zi = zi+γ. Suppose next that zi(t) 6= zi(t+1) for at least one value of t, and a, b have a common root c. Then there is a special letter zd in W , i ≡ i(t) ≡ d modulo µ or modulo ν (and hence in either case modulo γ).
Moreover for each t either zi(t) = zi(t+1) or each of zi(t), zi(t+1) is a power of c. Since zi(t) 6= zi(t+1) for at least one value of t, it follows that each zi(t) is a power of c. In particular zi and zi+γ are both powers of c.
Finally, suppose that zi(t) 6= zi(t+1) for at least one value of t, and a, b have no common root. By admissibility, a, b also have no common non-trivial power. As above, W ψ ψ contains a special letter zd = a or zd = b , and i ≡ d modulo γ. Consider the least t for which zi(t) 6= zi(t+1). Then either zi(t) and zi(t+1) are both powers of a or both powers of b. Assume the former. Then zi = zi(0) = zi(1) = ··· = zi(t) are also
45 Chapter 3: One-relator product induced from generalised triangle group
powers of a. We claim that zi(t+1),... zi(N) = zi+γ are also powers of a, which will complete the proof. Suppose by way of contradiction that this is not true. Then there is an m for which zi(m) is a power of a and zi(m+1) is not a power of a. By the definition of virtual periodicity, it follows that both zi(m) and zi(m+1) must be powers of b. But then zi(m) is simultaneously a power of a and of b, contrary to the admissibility hypothesis.
Corollary 3.5.5. If a clique-label has virtual period µ < l, then a refinement is possible.
Proof. By Lemma 3.5.4 the clique-label
n Y label(v) = [aα(j)Ubβ(j)U −1] (3.22) j=1 has virtual period gcd(µ, l)|l, so without loss of generality µ|l. Let V = z1 ··· zµ/2 and let s = l/µ. Then by definition of virtual periodicity and by the admissibility hypothesis one of the following is true:
1. s is odd and sn Y label(v) = [aγ(j)V bδ(j)V −1], (3.23) j=1 for some γ(j), δ(j).
2. s is even, a, b have a common root c, and
sn Y label(v) = [cγ(j)V xV −1], (3.24) j=1
for some letter x of order 2 and some γ(j).
In either case, we have a refinement.
Corollary 3.5.6. Suppose that v is a clique in a W-minimal clique-picture over G satisfying Hypothesis A. Suppose also that the generalised triangle group description of G has no refinement. Then the length of any zone incident at v is strictly less than l.
Proof. The i’th zone Zi contains ωi arcs. Assume that ωi ≥ l > l/2. Then by + Lemma 3.5.3, the word si has length ωi + µi − 1, which is strictly greater than + l + µi − gcd(l, µi) since gcd(l, µi) is even. Moreover, si is virtually periodic of + virtual period µi. But si is a cyclic subword of the clique-label label(v) which is virtually periodic with virtual period l.
46 Chapter 3: One-relator product induced from generalised triangle group
By Lemma 3.5.4 it follows that label(v) has virtual period gcd(l, µi) ≤ l/2. But by Corollary 3.5.5 this leads to a refinement of our generalised triangle group description of G, contrary to hypothesis.
3.6 The relator has free product length at least 4 and the pair (a, b) is admissible.
Assuming Hypothesis A, we have an W-minimal clique-picture over a one-relator product
(G ∗ G ) G = 1 2 , N(R)n with n ≥ 2 and `(R) ≥ 4 as a word in hai ∗ hUbU −1i. Any clique-label has the form
k Y label(u) = [aα(j)Ubβ(j)U −1] (3.25) j=1 By the Spelling Theorem 2.4.3 we must have k ≥ 2nl where l = `(aUbU −1). But by Corollary 3.5.6 each zone has fewer than l arcs, so all cliques have degree at least 2n + 1.
Theorem 3.6.1. Assuming Hypothesis A, any W-minimal clique-picture over G on D2 satisfies C(6).
To prove the Theorem 3.6.1, we assume that n = 2 and that v is a clique of degree
5, with zones Z1,... Z5 of sizes ω1, . . . ω5 respectively, in cyclic order around v, and aim to derive a contradiction. The key tool in the proof of Theorem 3.6.1 is the following.
Lemma 3.6.2. For each i = 1,..., 5, one of the following holds:
1. ωi + ωi−1 < 3l/2;
2. ωi + ωi+1 < 3l/2.
Proof. By Lemma 3.5.3, si is either an initial segment or a terminal segment of a + subword si of label(v) of length ωi +µi −1 and virtual period µi, where 0 < µi ≤ l/2. + We will assume that si is an initial segment of si and show that ωi + ωi+1 < 3l/2. + (An entirely analogous argument shows that, if si is a terminal segment of si , then
ωi−1 + ωi < 3l/2.)
47 Chapter 3: One-relator product induced from generalised triangle group
Now Lemma 3.5.3 to si+1: si+1 is either an initial segment or a terminal segment of + a cyclic subword si+1 of label(v) of length ωi+1 +µi+1 and virtual period µi+1, where
0 < µi+1 ≤ l/2. Our argument splits into two cases, depending on whether si+1 is + an initial or terminal segment of si+1. + Case 1. si+1 is a terminal segment of si+1.
Consider the cyclic subword W := sizs(i+1)si+1 of label(v). Since µi ≤ l/2 < ωi+1, + + the virtually periodic subword si is an initial segment of W . Similarly, si+1 is a terminal segment of W . These segments intersect in a segment of length
µi + µi+1 − 1 > µi + µi+1 − λ, (3.26)
where λ = gcd(µi, µi+1), so by Lemma 3.5.4 W is virtually periodic with period λ. Now recall that W is a cyclic subword of label(v), which is virtually periodic with virtual period l. If W has length greater than l + λ − 2, then label(v) has virtual period gcd(l, λ) < l, by another application of Lemma 3.5.4. But by Corollary 3.5.5 this leads to a refinement of our generalised triangle group description of G, contrary to hypothesis. Thus
ωi + ωi+1 = 1 + `(W ) ≤ l + λ − 1 < 3l/2. (3.27)
+ Case 2. si+1 is an initial segment of si+1. + Lets ¯i,s ¯i denote the cyclic subwords of label(v) that begin with the letter exactly l places after the first letter of si. By the virtual periodicity of label(v) it follows that + + s¯i is also virtually periodic, of virtual period µi. Moreover,s ¯i has length ωi +µi −1 and hass ¯i as an initial segment.
By construction, the union of the subwords si+1 ands ¯i of label(v) has length l − 1. + + Let W be the union of the subwords si+1 ands ¯i of label(v). Then W has length at least l + µi − 1. Arguing as in Case 1, we obtain a refinement of the generalised + + triangle group description of G, contrary to hypothesis, if si+1 ands ¯i intersect in a segment of length µi + µi+1 − gcd(µi, µi+1) or greater. So we may assume that this does not happen. + + In particular, µi+1 < l − ωi+1 + µi, for otherwises ¯i is a subword of si+1, of length
ωi + µi − 1 > l/2 + µi − 1 > µi+1 + µi − gcd(µi, µi+1). (3.28)
+ + + Hences ¯i is a terminal segment of W , and the intersection of si+1 ands ¯i has length precisely ωi+1 + µi+1 + ωi − l − 1. Thus
ωi + ωi+1 < µi + µi+1 − gcd(µi, µi+1) + l + 1 − µi+1 < l + µi ≤ 3l/2 (3.29)
48 Chapter 3: One-relator product induced from generalised triangle group as claimed.
Using Lemma 3.6.2, we complete the proof of Theorem 3.6.1 as follows. Renumbering the zones if necessary, we may assume by Lemma 3.6.2 that ω1+ω2 < 3l/2. Applying
Lemma 3.6.2 again, with i = 4, either ω3 + ω4 < 3l/2 or ω4 + ω5 < 3l/2. In the first case ω5 > l; in the second ω3 > l. Either of these is a contradiction.
3.7 The relator has no letter of order 2
In this section our main aim is to prove the following.
Theorem 3.7.1. Assuming Hypothesis B, any W-minimal clique-picture over G on D2 satisfies C(6).
In order to do so we will need a number of lemmas that are particular to the situation of Theorem 3.7.1, which we also collect together in this section. Recall that
(G ∗ G ) G = 1 2 , N(Rn) where n ≥ 2 and the relation R = R(a, UbU −1) contains no letters of order 2. We assume that v is a clique of degree less than 6 in a clique picture, joined to neighbouring cliques ui by zones Zi. The first step in our proof is designed to further restrict the form of R (and hence also of v).
Lemma 3.7.2. For any zone Zi, if s(i)+j ≡ t(i)+m modulo l where 1 ≤ j, m < ωi, then si has an element of order 2 in G1 or G2.
Proof. Suppose by way of contradiction that s(i) + j ≡ t(i) + m modulo l where
1 ≤ j, m < ωi. Recall that si is a cyclic subword of
n Y α(k) β(k) −1 label(v) = z0z1 ··· znl−1 = [a Ub U ]. (3.30) k=1
Write si = zs(i)+1 ··· zs(i+1)−1. Similarly, ti is a cyclic subword of
q Y γ(k) δ(k) −1 label(ui) = y0y1 ··· yql−1 = [a Ub U ]. (3.31) k=1
−1 0 Write ti = yt(i)+1 ··· yt(i+1)−1. By hypothesis, yt(i)+m = zs(i)+m0 for some m with 0 0 1 ≤ m < ωi. In particular s(i) + j ≡ t(i) + m ≡ s(i) + m modulo 2, since 0 yt(i)+m and zs(i)+m0 belong to the same free factor. Thus j + m is even. Moreover,
49 Chapter 3: One-relator product induced from generalised triangle group
−1 −1 zs(i)+(j+m0)/2 = yt(i)+(j+m0)/2. If zs(i)+(j+m0)/2 is a special letter, then so is yt(i)+(j+m0)/2. But in this case an amalgamation is possible, contrary to hypothesis. Otherwise zs(i)+(j+m0)/2 = yt(i)+(j+m0)/2, so zs(i)+(j+m0)/2 has order 2 in G1 or G2.
−1 Lemma 3.7.3. If no letter of R(a, UbU ) has order 2, then there is no zone Zi with ωi > l/2.
Proof. Suppose that ωi > l/2 for some zone Zi. Write si = zs(i)+1 ··· zs(i+1)−1 and ti = yt(i)+1 ··· yt(i+1)−1. By Lemma 3.7.2 we may assume that s(i) + j 6≡ t(i) + m modulo l for all j, m ∈ {1, 2, . . . , ωi − 1}, so in particular ωi ≤ l/2 + 1. Moreover, if
ωi = l/2 + 1 then we must have t(i) + 1 ≡ s(i + 1) modulo l. But since zs(i+1)−1 = −1 yt(i)+1 belongs to the same free factor as zt(i)+1 = zs(i+1), this is a contradiction.
Hence ωi ≤ l/2.
In what follows next, we will need the following result about generalised triangle groups.
Proposition 3.7.4. Let H = h x, y | xp, yq, (xy)2 i be a triangle group. If v(x, y) = xαyβxγyδ is trivial in H, with α, γ ∈ {1, 2, . . . , p − 1} and β, δ ∈ {1, 2, . . . , q − 1}, then one of the following holds:
1. 2 ∈ {p, q};
2. α = β = γ = δ = 1;
3. α = γ = p − 1 and β = δ = q − 1.
Proof. Assume that p 6= 2 6= q and consider the elements πi ! πi ! e p 0 e q t X = −πi and Y = −πi in PSL2(C). It follows that X and Y 1 e p 0 e q π π have orders p and q respectively in PSL2(C). If we take t = −2 cos p + q , then T r(XY ) = 0 and hence the map x 7→ X, y 7→ Y extends to a faithful representation α β −δ −γ of H in PSL2(C). Suppose that X Y = Y X . By comparing the left lower entries of both sides of the equation we have α ≡ ±γ modulo p and
απ βπi γπ δπi sin e q ∓ sin e q = 0. (3.32) p p
By expanding and solving component-wise, we have that
γπ β − δ sin sin π = 0. (3.33) p q
In particular, β = δ. Similarly we have α = γ.
50 Chapter 3: One-relator product induced from generalised triangle group
Hence v = (XαY β)2 = ±I. By comparing off-diagonal entries, we conclude that Xα 6= ±Y −β, so v 6= +I. Hence v = −I, and so T r(XαY β) = 0. In other words
απ βπ α β sin p sin q 2 cos + π + t π π = 0. (3.34) p q sin p sin q
Hence we deduce that απ βπ π π tan tan = tan tan . (3.35) p q p q Since p, q > 2, the last equality holds if and only if either α = β = 1 or α = p − 1 and β = q − 1.
The clique v fails to satisfy the C(6) property, so by Lemma 3.7.3 its label has length at most 5l/2. But this length is a multiple of l, so at most 2l. By Theorem 2.4.3 and Proposition 3.7.4 we can assume that R = aUbU −1 (up to cyclic permutation) and that the label of v is label(v) = (aUbU −1)±2 (up to cyclic permutation). Without loss of generality we will assume throughout that this label is label(v) = (aUbU −1)2 and the letters a, b will mean the corresponding special letters. This remark enables us to strengthen Lemma 3.7.3 as follows.
−1 Lemma 3.7.5. If no letter of R(a, UbU ) has order 2, then there is no zone Zi with ωi ≥ l/2.
Proof. By Lemma 3.7.3 we are reduced to the case where ωi = l/2. By Lemma 3.7.2, we must have t(i) ∈ {s(i+1)−1, s(i+1), s(i+1)+1} modulo l. The first possibility leads to a contradiction as in the proof of Lemma 3.7.3. The third possibility also leads to a contradiction for similar reasons, since it implies that t(i + 1) ≡ s(i) + 1 modulo l. Hence we may assume that t(i) ≡ s(i + 1) modulo l and hence that
s(i) ≡ s(i + 1) + l/2 ≡ t(i) + l/2 ≡ t(i + 1) modulo l. (3.36)
If zs(i) is special, then so is yt(i+1), and we may amalgamate cliques, contrary to hypothesis. Hence zs(i) is not special. In other words s(i) 6≡ 0 modulo l/2. Thus si contains precisely one special letter – say a without loss of generality. Hence also t contains precisely one special letter, which is necessarily a power of b. Thus ψ −1 −1 aUb U is a proper cyclic permutation of zs(i)sizs(i+1)ti ≡ zs(i)sizs(i+1)si for some ψ. Applying Lemma 3.2.7, we see that
s Y aUbψU −1 = [aα(k)V dβ(k)V −1] (3.37) k=1
ψ for some word V , some α(k), β(k) = ±1 and some d ∈ {b , zs(i), zs(i+1)}, where s > 1.
51 Chapter 3: One-relator product induced from generalised triangle group
Moreover, from the proof of Lemma 3.2.7 we see that we may take d = bψ if s is odd, while if s is even then a = bψ. In either case, hai ∗ hUbU −1i is a proper subgroup of hai ∗ hV bV −1i or hai ∗ hV dV −1i, giving a refinement of the generalised triangle group description of G. This contradiction completes the proof.
Remark 3.7.6. From Lemma 3.7.5, our interior clique v which fails the C(6) condition must have exactly five zones. Let these zones be Z1,Z2,...,Z5 listed consecutively in clockwise order.
Proposition 3.7.7. There are exactly three or four of the zones Z1,Z2,...,Z5 containing a special letter.
Proof. By assumption
−1 2 label(v) = (aUbU ) = z0z1 ··· z2l−1 (3.38)
so there are precisely four special letters in label(v), namely z0 = zl = a and zl/2 = z3l/2 = b. By Lemma 3.7.3 no zone can contain more than one special letter, so it suffices to show that at most one of the special letters is not contained in a zone. Suppose by way of contradiction that z0 and zl are not contained in zones. By
Lemma 3.7.5, each of the subwords z1 ··· zl−1 and zl+1 ··· z2l−1 must contain at least three zones, contradicting our assumption that there are only five zones in total. A similar contradiction arises if z0 and zl/2 are not contained in zones: z1 ··· zl/2−1 and zl/2+1 ··· z2l−1 must contain at least two and four zones respectively. By symmetry, any other combination of two special letters not contained in zones also contradicts our underlying hypotheses, hence the result.
Therefore we have exactly two possible configurations as depicted in the Figures 3.12 and 3.13 below.
Figure 3.12: Diagram showing the case where only the special letters z0, zl/2 and zl are contained in zones
52 Chapter 3: One-relator product induced from generalised triangle group
Figure 3.13: Diagram showing the case where all four special letters z0, zl/2, zl and z3l/2 are contained in zones
Remark 3.7.8. For each zs(i), we have s(i) 6= s(j) modulo l/2 for i 6= j since no
ωi ≥ l/2. Remark 3.7.8 gives the inequalities
0 < s(3) < s(1) < l/2 = s(4) < s(2) < s(5) ≤ l − 1 (3.39) and 0 < s(3) < s(1) < l/2 < s(4) < s(2) < s(5) ≤ l − 1 (3.40) corresponding to Figure 3.12 and Figure 3.13 respectively. (Note that s(i) in the inequalities is the modulo l equivalent. The actual value can be read from the Figures 3.12 and 3.13. For example the actual value for s(4) in Figure 3.12 is 3l/2.) The corresponding subwords are:
s1 = zs(1)+1 . . . zs(2)−1,
s2 = zs(2)+1 . . . zs(3)−1,
s3 = zs(3)+1 . . . zs(4)−1,
s4 = zs(4)+1 . . . zs(5)−1,
s5 = zs(5)+1 . . . zs(1)−1.
Lemma 3.7.9. Let W be a cyclically reduced word of length 2m in a free product. Let X be a reduced word of length m such that both X and X−1 appear as cyclic subwords of W . Then X contains a letter of order 2.
Proof. The subwords X and X−1 of W must intersect, for otherwise W is a cyclic conjugate of XX−1, contradicting the fact that it is cyclically reduced. Hence there
53 Chapter 3: One-relator product induced from generalised triangle group is an initial segment Y of X or X−1 that coincides with a terminal segment of X−1 or X respectively. Thus Y ≡ Y −1 and so Y has an odd length and its middle letter has order 2.
More notations
−1 Think of aUbU = z0 ··· zl−1 as a cyclic word satisfying a partial reflectional sym- metry using the special letters as mirrors. Thus U has mirror image U −1. More generally the mirror image of zj . . . zk is zl+2−j . . . zl+2−k ( subscripts modulo l ). Unless stated otherwise, X (with or without subscript) denotes an initial or termi- nal segment of U. Similarly Y (with or without subscript) will denote an initial or terminal segment of U −1. Also X−1 and Y −1 are mirror images of X and Y respectively. Using this notation, we can express the subwords in the zones as:
s1 = X1x1Y1,
s2 = Y2y2X2,
s3 = X3x3Y3,
s5 = Y5y5X5.
where xi = b and yi = a are the corresponding special letters. In other words if Xi is a terminal segment of si ( as in the case of s2), then it is an initial segment of U. In −1 which case Yi is a terminal segment of U . If s(4) = l/2, then s4 = Y4. Otherwise, −1 s4 is a subword of U which is neither an initial nor terminal segment. Note that some of the xi,Xi, yi and Yi are allowed to be empty as in the case of s3 in Figure 3.13.
0 0 Let si be the mirror image of ti. Then si is identically equal to si (modulo ∼). In 0 0 0 0 0 0 0 particular if si = Xix1Yi, then si = Xix1Yi where Xi ∼ Xi, x1 ∼ x1 and Yi ∼ Yi. We use M to denote an initial segment of U or a terminal segment of U −1. Similarly −1 + N denotes a terminal segment of U or an initial segment of U . Mk is the initial segment of U or terminal a terminal segment of U −1 of length `(M) + k. If M is a 0 subword of si, then Mi denotes the image of M under Zi and M is the mirror image + 0 of Mi. Nk , Nj and N are defined similarly. Remark 3.7.10. One thing to note is that if M is a subword of U, say, with `(M) ≥ 0 l/4, then Mi can not be a subword of U by Lemma 3.7.2. So that either M is a subword of U or neither a subword of U nor U −1. In the case where M and M 0 −1 + are both subwords of U or U , then what we call Mk will be the union of the + 0 0 two. It follows that `(Mk ) > `(M) for otherwise M = M (i.e M is also an initial
54 Chapter 3: One-relator product induced from generalised triangle group
−1 segment of U), and hence Mi is a terminal segment of U , and hence there is an amalgamation of cliques, contrary to hypothesis.
Lemma 3.7.11. Let M, M 0 and `(M)+k be as in Remark 3.7.10. If `(M) ≥ l/4 and 0 −1 + + both M and M are subwords of U or U , then Mk has period γ = `(Mk )−`(M) ≤ `(U) − `(M).
The proof of Lemma 3.7.11 follows easily from Remarks 3.2.4 and 3.7.10.
Lemma 3.7.12. Suppose that U has a period γ < `(U) with X and Y as initial and terminal segments respectively both of length γ/2. Then no segment of si is of the −1 −1 form XyiX or Y xiY .
−1 Proof. Suppose without loss of generality that si has XyiX as a segment. Then it is identically equal to a subword of UxU −1. Thus X is identically equal to a subword of U or U −1. Take W to be any subword of U of length γ. The periodicity of U implies that each of X,X−1 is identically equal to a cyclic subword of W . By Lemma 3.7.9 it follows that X contains a letter of order 2 contrary to hypothesis.
Lemma 3.7.13. Suppose that W has period γ < `(W ) and no element of order 2. Then W has no subword of the form L = wrw−1 with `(w) ≥ γ/2.
Proof. Suppose by contradiction that W has a subword of the form L = wrw−1 with `(w) ≥ γ/2. Let r1 and r2 be the letters γ/2 places before and after r in L −1 respectively, then r1 = r2. But by the periodicity of W , r1 = r2. Hence W has an element of order 2 contradicting hypothesis.
Lemma 3.7.14. Let X be an initial (respectively terminal) segment of U of length
`(X) ≥ l/4. Suppose si = Y aX (respectively si = XaY ) for some terminal (respec- −1 0 tively initial) segment Y of U . If si is not contained entirely in U, then U has period γ ≤ 2(`(U) − `(X)). Furthermore if γ = 2(`(U) − `(X)), then U has a termi- −1 −1 nal (respectively initial) segment of the form xw zs(i+1)w (respectively wzs(i)w x) for some letter x and some word w with `(w) = γ/2 − 1.
Proof. By symmetry it suffices to prove the first case, where X is an initial segment of U. Write U ≡ Xzs(i+1)V for some terminal segment V of U. Consider the mirror 0 ψ −1 0 image si of ti. By Lemma 3.7.2, ti is a subword of V b U for some ψ. Hence si −ψ −1 0 −ψ −1 is a subword of Ub V . Therefore si has the form w1b w2 for some terminal l segments w1 and w2 of U with `(w2) ≤ `(V ) = `(U) − (`(X) + 1) < 4 − 1. Denote by S the terminal segment of U of length
55 Chapter 3: One-relator product induced from generalised triangle group
`(S) = `(X) − `(w2) − 1
0 = `(si) − `(w2) − `(Y ) − 2
= `(w1) − `(Y ) − 1. (3.41)
−ψ −1 Then w1 ≡ Y aS and X ≡ Sb w2 and so S is identically equal to an initial segment of U. Thus by Remark 3.2.4, we have that U has period
γ = `(U) − `(S)
= `(U) − `(X) + `(w2) + 1
≤ 2(`(U) − `(X)). (3.42)
Finally if γ = 2(`(U) − `(X)), then 1 + `(X) + `(w2) = `(U). Thus V ≡ w2 and −ψ −1 so U ≡ Xzs(i+1)V ≡ Sb w2 zs(i+1)w2. The result follows by talking w = w2 and x = b−ψ.
Lemma 3.7.15. Let Zi and Zj be distinct. Suppose that the intersection of si and −1 sj contains a subword of the form L = wφw where φ is a special letter. Let Li and
Lj be the images of L under Zi and Zj respectively, and [Li], [Lj] their ∼-classes in ∗ (Ω/ ∼) . Then [Li] and [Lj] do not intersect as cyclic subwords of the ∼-class of aUbU −1 in (Ω/ ∼)∗.
Proof. Let σ be the intersection of si and sj. Since σ is non-empty, the zones Zi and Zj are not consecutive, so i − j ≡ ±2 modulo 5. Without loss of generality, suppose that j ≡ i + 2 modulo 5. Then σ is an initial segment of si and a terminal segment of sj. So σ = zs(i)+1 . . . zs(j+1)−1 where sj = zs(j)+1 . . . zs(j+1)−1 and si = zs(i)+1 . . . zs(j−1)−1. Choose L with maximal length among all subwords of σ of that form. Let % be the union of si and sj. Then % = zs(j)+1 . . . zs(j−1)−1 and a cyclic 2 permutation of the clique-label of v has the form (%zs(j−1)sj−1zs(j)) .
0 0 0 Now define tj = zt(j)+1 . . . zt(j+1)−1 and ti = zt(i)+1 . . . zt(i+1)−1. Then ti is identical to the image ti of si, with the possible exception of a special letter of ti: the cor- 0 responding letter of ti is also special, and these two special letters may be different 0 powers of a (or of b). Similarly, tj agrees with tj except possibly at a special letter. 0 Now define σi to be the terminal segment σi = zd+1 . . . zt(i+1)−1 of ti, where
d := t(i + 1) + s(i) − s(j + 1) modulo l, (3.43)
56 Chapter 3: One-relator product induced from generalised triangle group
0 and σj to be the initial segment σj = zt(j)+1 . . . ze−1 of tj, where
e := t(j) + s(j + 1) − s(i) modulo l. (3.44)
Then σi and σj agree with the images Li and Lj of σ under Zi and Zj respectively, again with the possible exception that they may differ at a special letter.
−1 If [Li] and [Lj] coincide as cyclic subwords of [aUbU ], then σi and σj coincide −1 0 0 as cyclic subwords of aUbU . In this case we can form the union σ∗ = ti ∪ tj = −1 zt(i)+1 . . . zt(m)−1: a cyclic subword of aUbU that is disjoint from σ and hence contains at most one special letter. But σ∗ is identically equal to %−1, with the possible exception of a special letter of σ∗. It follows from Corollary 3.7.5 that `(%) > l/2. Thus % ∩ σ∗ 6= ∅ and so without loss of generality some initial segment τ of % coincides with a terminal segment of σ∗. Since the only special letter of % is φ which does not appear in σ∗, τ does not contain a special letter. It follows that τ −1 ≡ τ, whence τ contains a letter of order 2, contrary to hypothesis.
Suppose then that [Li] and [Lj] intersect but do not coincide. Consider the subword 0 0 ∗ −1 %∗ = σi ∪ σj of ti ∪ tj. As above, % is a cyclic subword of aUbU which is disjoint from σ and hence contains at most one special letter. Write %∗ = x1x2 . . . xr for some r. Note that r is odd, since by definition σ begins and ends with letters in the same free factor, and hence the same holds for %.
Assume without loss of generality that σi, σj are the initial and terminal segments respectively of %∗. Then
σi = x1x2 . . . x`(L) (3.45) and
σj = xr+1−`(L)xr+2−`(L) . . . xr. (3.46)
−1 ∗ Since each of σi, σj agrees with σ except possibly at a special letter of % , it follows that, for 1 ≤ µ ≤ `(L),
xµ = xr+µ−`(L) (3.47)
∗ unless one of xµ, xr+µ−`(L) is a special letter of % . Also, since σ−1 agrees with σ except for the middle letter φ±1, it follows that for 1 ≤ µ ≤ `(L), −1 xµ = x`(L)+1−µ (3.48)
−1 unless either µ = (`(L) + 1)/2 or one of xµ = x`(L)+1−µ is a special letter of %. Similarly, for r + 1 − `(L) ≤ µ ≤ r,
−1 xµ = x2r−`(L)−µ (3.49)
57 Chapter 3: One-relator product induced from generalised triangle group
−1 ∗ unless either µ = r − `(L)/2 or one of xµ = x2r−`(L)−µ is a special letter of % . ∗ Now consider the three letters x`(L)+1−(r+1)/2, x(r+1)/2 and x3(r+1)/2−`(L)−1 of % . We know at most one of these three letters can be special.
Suppose first that neither x`(L)+1−(r+1)/2 nor x(r+1)/2 is special. Then
−1 x r+1 = x r+1 (by Equation 3.48) 2 `(L)+1− 2
−1 = x r+1 (by Equation 3.47) r+`(L)+1− 2 −`(L)
−1 = x r+1 (3.50) 2
It follows that x(r+1)/2 has order 2, contrary to hypothesis.
Similarly, if neither x(r+1)/2 nor x3(r+1)/2−`(L)−1 is special, then we may deduce that x(r+1)/2 using Equations 3.47 and 3.49.
∗ −1 Finally, suppose that x(r+1)/2 is a special letter of % . Then xµ = xr+1−µ for each µ = 1, 2,..., (r − 1)/2. In particular
−1 x `(L)+1 = x `(L)+1 . (3.51) 2 r+1− 2
But by Equation 3.47,
x `(L)+1 = x `(L)+1 . (3.52) 2 r+1− 2
Hence x(`(L)+1)/2 has order 2, again contrary to hypothesis. This completes the proof.
Remark 3.7.16. We remark that the first part of the proof of Lemma 3.7.15 does not assume any form for the intersection.
3.7.1 Proof of Theorem 3.7.1
We are now ready to complete the proof of Theorem 3.7.1. Our proof is by contra- diction. Suppose that some interior vertex v of Γ fails to satisfy C(6), then we know from Remark 3.7.6 that v has exactly five incident zones Z1,. . . , Z5. We also assume by Corollary 3.7.5 and Remark 3.7.6 that ωi < l/2 for i = 1,..., 5. The proof is divided into two cases. Case 1. s(4) = l/2. Case 2. s(4) > l/2.
58 Chapter 3: One-relator product induced from generalised triangle group
Proof of Case 1. For Case 1 (alternatively case 3.39), take N to be the longer of X3 and Y4 and M to be the longer of X5 and Y2. In each case M is either an initial segment of U or a terminal segment of U −1. Similarly N is either a terminal segment −1 of U or an initial segment of U . Also `(N), `(M) ≥ 1/4 for otherwise `(si) ≥ l/2 for some i ∈ {1, 2, 5}, contradicting hypothesis. By Lemma 3.7.14, U has an initial segment of period ρ ≤ 2(`(U) − `(M)) and a terminal segment of period γ ≤ 2(`(U) − `(N)). If γ < 2(`(U) − `(N)), then
L = zl−γ/2zl+1−γ/2 . . . zl+γ/2 is a proper subword of s2 ∩ s5. It follows from Lemma
3.7.15 that at least one of the images of L under Z2 and Z5 is identically equal to a periodic subword of U (with period γ). This can not happen by Lemma 3.7.13. Otherwise U has period γ = 2(`(U) − `(N)). It is easy to see that
`(X1), `(Y1) ≥ `(U) − `(N) (3.53)
as otherwise either `(s2) ≥ l/2 or `(s5) ≥ l/2, thereby contradicting hypothesis.
Hence the result follows by applying Lemma 3.7.12 to s1.
For Case 2 (alternatively case 3.40), take N to be the longer of X3 and Y1 and M to be the longer of X5 and Y2. In each case M is either an initial segment of U or a terminal segment of U −1. Similarly N is either a terminal segment of U or an initial segment of U −1. Also `(N), `(M) ≥ 1/4. The proof is subdivided into three sub-cases namely: Case 2a. Each of M 0 or N 0 is identically equal to a subwords of U −1 or U. Case 2b. Exactly one of M 0 or N 0 is identically equal to a subword of U or U −1. Case 2c. Both M 0 and N 0 are not identically equal to a subword of U −1 or U.
Proof of Case 2a. Using Lemma 3.7.11, we conclude that U has a periodic initial + + segment Mγ of period γ ≤ `(U) − `(M) and a periodic terminal segment Nρ of period ρ ≤ `(U) − `(N). Moreover by Remark 3.7.6, the two segments intersect in a segment S with
+ + `(S) = `(Mγ ) + `(Nρ ) − `(U) ≥ ρ + γ + 1 > ρ + γ.
Hence U is periodic with period ν = gcd(ρ, γ) ≤ min{ρ, γ} ≤ min{`(Y1), `(X1)} −1 by Corollary 3.2.6. It follows that s1 contains a subword the form wx1w with `(w) = ν/2. The result will then follow from Lemma 3.7.12.
59 Chapter 3: One-relator product induced from generalised triangle group
Proof of Case 2b. Suppose N 0 is identically equal to a subword of U or U −1. By Lemma 3.7.14, U is periodic with period ν ≤ 2(`(U)−`(M)). If ν < 2(`(U)−`(M)), the result follows by applying Lemma 3.7.12 to s1 as in Case 2a. Hence we may assume that ν = 2(`(U) − `(M)). Then by Lemma 3.7.14 U a has −1 a terminal segment of the form u1wu2w for some letters u1, u2 and some word w with `(w) = ν/2 − 1. + + Also by Lemma 3.7.11, U has a periodic terminal segment Nρ of length `(Nρ ) = + `(N) + ρ and period ρ ≤ `(U) − `(N). If ρ + 2 ≤ ν, then Nρ has a subword of the −1 formwu ˆ 2wˆ wherew ˆ is the terminal segment of w of length ρ/2. This contradicts Lemma 3.7.13. Finally if ρ + 2 > ν, then ρ + ν ≤ 2ρ + 1 ≤ `(N) + ρ. It follows from Corollary 3.2.6 that U has period λ = gcd(ν, ρ). Hence we can apply Lemma 3.7.12 to s2 since
λ ≤ λ − 1 ≤ ρ − 1 ≤ `(U) − max{`(X ), `(Y )} − 1 ≤ min{X ,Y }. (3.54) 2 3 1 2 2
The proof for the case where M 0 is identically equal to a subword of U or U −1 is similar.
Proof of Case 2c. Then U is periodic with periods ρ ≤ 2(`(U) − `(M)) and γ ≤ 2(`(U) − `(N)). If ρ < 2(`(U) − `(M)) or γ < ρ, then the result follows by applying
Lemma 3.7.12 to s2. Similarly if γ < 2(`(U)−`(N)) or ρ < γ, then the result follows by applying Lemma 3.7.12 to s1.
Hence suppose ρ = γ = 2(`(U) − `(M)) = 2(`(U) − `(N)). Then we have `(X5) ≤
`(M) = `(U) − γ/2, whence `(X1) = `(U) − 1 − `(X5) ≥ γ/2 − 1 and so
`(Y1) = `(s1) − 1 − `(X1) < `(U) − 1 − `(X1) ≤ `(U) − γ/2 = `(N). (3.55)
It follows that N = X3. By a similar argument, M = X5. It follows from Lemma −1 3.7.14 that U has an initial segment of the form w1zs(3)w1 u1 and a terminal segment −1 −1 of the form u2w2zs(1)w2 for some letters u1 = zρ, u2 = zl/2−ρ, and words w1, w2 satisfying `(w1) = `(w2) = ρ/2 − 1.
We also have from Lemma 3.7.14 that t(3) + ω3 ≡ s(3) ≡ ρ/2 modulo l. The
(l − ρ)/2-th letter of s3 is x3 = zl/2 = b. The corresponding letter of t3 is therefore −1 −1 zs(3)−(l−ρ)/2 = zl/2+ρ, which is not a special letter of t3. Hence zl/2+ρ = zl/2 = b . Therefore −1 u2 = z l −ρ = z l = b. (3.56) 2 2 +ρ A similar argument shows that
−1 u1 = zρ = zl−ρ = y5 = a. (3.57)
60 Chapter 3: One-relator product induced from generalised triangle group
Moreover by the periodicity of U
−1 w1zs(3)w1 a = z1 ··· zρ (3.58) is a cyclic conjugate of −1 −1 bw2z w2 = z l ··· z l . (3.59) s(1) 2 −ρ 2 −1 Now apply Lemma 3.2.7 to this pair of cyclically conjugate words to obtain
s −1 −1 Y β(t) α(t) −1 bw2zs(1)w2 = b V x V (3.60) t=1 for some letter x, some word V and some integer s and some α(t), β(t) ∈ {±1}. The integer s in the statement of Lemma 3.2.7 is defined to be k/m, where k = ρ/2 and m = gcd(j, k). Here j in turn is defined as the number of places by which one has −1 −1 −1 to cyclically permute bw2zs(1)w2 to obtain w1zs(3)w1 a – in other words
z1 ··· zρ ≡ z l ··· z l z l ··· z l . (3.61) 2 −ρ+j 2 −1 2 −ρ 2 −ρ+j−1
−1 Suppose first that s is even. Then Lemma 3.2.7 gives that b = zs(1), a = zs(3), and x = a in the above expression. This U has a terminal segment
s −1 −1 Y β(t) α(t) −1 bw2zs(1)w2 = b V a V (3.62) t=1
Similarly U has an initial segment of the form
s −1 Y −1 δ(t) γ(t) w1zs(3)w1 a = V b V a (3.63) t=1
Putting these together, using the periodicity of U, we obtain
p Y U = V −1 bζ(t)V aη(t)V −1 (3.64) t=1 for some integer p and some ζ(t), η(t) ∈ {±1}. Thus UbU −1 ∈ ha, V −1bV i and we have a refinement of our generalised triangle group description of G, contrary to our underlying hypotheses.
−1 Next suppose that s is odd in Lemma 3.2.7. Then {b, zs(1)} = {a, zs(3)}, and we have an expression s −1 −1 Y β(t) α(t) −1 bw2zs(1)w2 = b V zs(1) V (3.65) t=1
61 Chapter 3: One-relator product induced from generalised triangle group and an analogous expression
s −1 Y −1 δ(t) γ(t) w1zs(3)w1 a = V zs(3)V a (3.66) t=1
Again we can fit these together using the periodicity of U to get an expression for U. −1 If b = zs(3) and a = zs(1) then again this has the form
p Y U = V −1 bζ(t)V aη(t)V −1 (3.67) t=1 which leads to a refinement, contrary to hypothesis. −1 If on the other hand b = a and zs(3) = zs(1) then we obtain an expression of the form
p η(0) −1 Y ζ(t) η(t) −1 −1 U = V zs(3) V a V zs(3)V ∈ ha, V zs(3)V i (3.68) t=1
As before, this yields a refinement, contrary to hypothesis. This completes the proof.
3.8 Proof of Theorems
Here we give the proofs for Theorems 3.1.1, 3.1.2 and 3.1.3. As before we suppose that the triangle group description for G is maximal. By Theorems 3.6.1 and 3.7.1, an W-minimal clique-picture over G satisfies the C(6) property.
Theorem 3.8.1 (Theorem 3.1.1). The maps G1 → G, G2 → G and H → G are all injective.
Proof. Suppose that there is a non-trivial word w in H, G1 or G2 that is trivial in G. Then we obtain a W-minimal picture P over G on D2 with boundary label w. We prove the theorem by induction on the number of cliques in P ; the case of 0 cliques corresponds to the empty picture P , for which there is nothing to prove. Suppose first that some clique v in P is not simply connected, and let C be one of the boundary components of the surface carrying the clique v. By an innermost argument, we may assume that C bounds a disc D ⊂ D2 such that every clique in P ∩ D is simply-connected (see Figure.
62 Chapter 3: One-relator product induced from generalised triangle group
Figure 3.14: Diagram showing a clique containing vertices v1, v2, . . . , v5 which is not simply-connected. The region bounded by v1, v5, v4, v5 and containing v6 (indicated by the red curve) contains only simply-connected cliques, and so is a good choice for C since v6 is in P ∩ C.
Now the label on C is a word in H which is the identity in G, and P ∩ D has at least one fewer clique than P , so by inductive hypothesis the label is trivial in H. We then amend P by replacing P ∩ D by a picture over H, all of the arcs, vertices and regions of which will belong to the same clique as v in the amended picture P 0. Since C ∩ P was not empty (for otherwise D is contained in v), the new picture P 0 also has fewer cliques than P , and the result follows from the inductive hypothesis. Hence we are reduced to the situation where every clique in P is simply connected, and hence we may form from P a clique-picture Γ over G on D2 with boundary label w. Without loss of generality, we may assume that Γ is W-minimal. It follows that Γ satisfies C(6).
If Γ is empty, then w is already trivial in H, G1 or G2 and so we get a contradiction. On the other hand suppose that Γ is non-empty. If no arcs of Γ meet ∂D2, then Γ is a spherical picture (i.e a picture on S2) and the C(6) property implies χ(S2) 6= 2.
This contradiction implies G1 → G and G2 → G are both injective. 2 Suppose then that some arcs of Γ meet ∂D . Then w 6∈ G1,G2. Moreover if w is a word in {a, UbU −1}, then the C(6) condition combined with Theorem 1.5.5 implies that some boundary clique v0 has at most degree three. 2 Under Hypothesis A, by Lemma 3.5.6, v0 is connected to ∂D by a zone Zi with
ωi > l. Either a refinement is possible by Corollary 3.3.1 or we can amalgamate 2 v0 with ∂D to form a new clique-picture with fewer cliques whose boundary label also belongs to H. The former possibility contradicts the maximality of the triangle group description for G while the latter contradicts the minimality of Γ.
2 Under Hypothesis B, it follows from Remark 3.7.6 that v0 is connected to ∂D by a zone Zi with ωi > l/2. By Corollary 3.7.5, U has a letter of order 2 or a refinement is possible, contradicting hypothesis. Otherwise as before we can amalgamate v0 with ∂D2 to form a new clique-picture with fewer cliques. Either case is a contradiction.
63 Chapter 3: One-relator product induced from generalised triangle group
Theorem 3.8.2 (Theorem 3.1.2). If the word problems are soluble for H, G1 and
G2, then it is soluble for G.
Proof. Any clique-picture Γ over G satisfies C(6), and hence a quadratic isoperi- metric inequality by Theorem 1.5.6. In other words, there is a quadratic function f such that any word of length m representing the identity element of G is the boundary label of a clique-picture with at most f(m) cliques. Also there is a bound (as a function of m) on the length of any clique-label of Γ. By Corollary 3.5.6 and Corollary 3.7.5, a clique with label of length ml has degree at least m. Moreover, there is a linear isoperimetric inequality of the form
X `(Γ) ≥ [deg(v) − 6] (3.69) v where `(Γ) is the number of regions of Γ and the sum is over all cliques v. Hence no clique can have degree greater than `(Γ) + 6f(`(Γ)). Since no zone has length greater than l, this gives an upper bound of l[`(Γ) + 6f(`(Γ))] on the length of any clique-label. Since both the number of cliques and the length of any clique-label are bounded, there are only a finite number of connected graphs that could arise as clique-pictures for words of length less than or equal to that of a given word w. Moreover, any such graph can be labelled as a clique-picture only in a finite number of ways. For any such potential labelling, we may check whether or not the clique-labels are equal to the identity in H, and whether or not the region-labels are equal to the identity in G1 or G2, using the solution to the word problem in H, G1 and G2 respectively. Hence we may obtain an effective list of all words of length less than or equal to `(w) that appear as boundary labels of connected clique-pictures over G. In particular, we may check, for all cyclic subwords w1 of w, whether or not w1g belongs to this list for some letter g ∈ G1 ∪ G2. (Note that this check also uses the solution to the word problem in G1 and G2, and that the letter g, if it exists, is unique by the Freiheitssatz). If so, then w is a cyclic conjugate of w1w2 for some w2 , so w = 1 in G if and only if g = w2 in G, which we may assume inductively is decidable. Hence the word problem is soluble for G.
n Theorem 3.8.3 (Theorem 3.1.3). If a cyclic permutation of R has the form W1W2 n with 0 < `(W1), `(W2) < `(R ) as words in G, then W1 6= 1 6= W2 as words in G. In particular R has order n in G.
Proof. Suppose by way of contradiction that W1 = 1 = W2 in G. We can assume by Theorem 3.1.1 that `(W1) 6= 1 6= `(W2). We obtain a W-minimal clique-picture
Γ over G with boundary label W1 or W2. Suppose without loss of generality that Γ ˜ −1 has boundary label W1. Form a new clique picture Γ with boundary label W2 by
64 Chapter 3: One-relator product induced from generalised triangle group adding a vertex labelled R−n. Γ˜ has only one boundary vertex and is reduced since Γ is W-minimal. It follows from Theorem 1.5.5 that Γ˜ has a single vertex or clique.
Hence up to conjugacy W2 (and hence W1) is a word in H which is trivial and has length strictly less than the length of Rn. This contradicts the Spelling Theorem, hence the result.
Theorem 3.8.4. [12] The pushout of groups in Figure 3.1 is geometrically Mayer- Vietoris in the sense of [23]. In particular it gives rise to Mayer-Vietoris sequences
· · · → Hk+1(G, M) → Hk(A ∗ B,M) →
Hk(G1 ∗ G2,M) ⊕ Hk(H,M) → Hk(G, M) → · · · and k k k · · · → H (G, M) → H (G1 ∗ G2,M) ⊕ H (H,M)
→ Hk(A ∗ B,M) → Hk+1(G, M) → · · · for any ZG-module M.
The proof of Theorem 3.8.4 is done by constructing a push-out square of aspherical CW- complexes and embeddings which realises Figure 3.1 on fundamental groups. These ideas are not covered in this thesis, so the reader can see the given reference for a complete proof. Also the proof uses the C(6) property of clique-pictures and Theorem 3.1.1, which justifies its inclusion here.
65 Chapter 4
One-relator product of two non-trivial groups with short relator
4.1 Preamble
Let A and B be non-trivial groups. Consider the one-relator product
(A ∗ B) G = , (4.1) N(rn) where n is a natural number and r is a word in A ∗ B which is not a proper power. We will be mostly interested in the case where n ≥ 2 and `(r) ≤ 8. Our aim is to show that G satisfies the Freiheitssatz and r has order n in G. In very few cases it might not be possible to prove both results. However, either of the two results will be enough for later applications. More precisely using pictures and clique pictures, we prove the following theorem.
Theorem 4.1.1. Let A and B non-trivial groups and G = (A ∗ B)/N(rn). Suppose that any of the following conditions holds:
1. `(r) = 8 and n ≥ 3; or
2. 2 ≤ `(r) < 8 and n ≥ 2; or
3. n = 1 and `(r) = 4 say r = abcd, with ha, ci and hb, di isomorphic to 2-
generator subgroups of PSL2(C).
Either:
i. A and B embed in G naturally and r has order n in G, or
66 Chapter 4: One-relator product of two non-trivial groups with short relator
ii. n = 2 and up to conjugation r has the form r = axbx−1cz with z2 = 1, where a, b, c ∈ A and x, z ∈ B.
Corollary 4.1.2. Let G be as in Theorem 4.1.1. Then r has order n in G.
Proof. Suppose that n = 2 and the relations xy = z2 = 1 holds in B, then one can show easily that B embeds in G. More precisely the identity map from B to itself factors through the projection map from G to B. So if r fails to have order 2, then r = 1 in G and so the image of r (under this projection map) which is z is trivial in B. But this is impossible. This completes the proof.
The rest of this Chapter is divided into five sections. In each section we consider a specific `(r) = l. But first we make some general assumptions as well as prove general results which apply to arbitrary l.
General assumptions
Most of the proofs we present in this chapter dwells on analysis of pictures. These pictures have some properties which we shall establish. We show by rigorous analysis that these properties will give us the desired result. If the group G in 4.1 fails to satisfy the Freiheitssatz or the word r has order m strictly less than n, then we can construct a non-trivial picture M over G on D2. In the first case M has an exceptional region ∆o whose label is a non-trivial word in A ∪ B which is trivial in
G. In the second case M has an exceptional vertex vo whose label is a non-trivial word r±m in A ∗ B which is trivial in G. In both cases M can in fact be thought of as a spherical picture. Also we will always assume that M is W-minimal, where W is the union of the set on non-trivial elements in A ∪ B and the symmetrized closure of the set {rk | k < n}.
If both A and B have faithful representations in PSL2(C) and n > 1, then G satisfies the Freiheitssatz and r has order n in G by Theorem 2.4.2. Hence we assume that at least one of A or B does not have any faithful representation in PSL2(C). In particular if A – say, is cyclic or dihedral, then B is neither cyclic nor dihedral. Another assumption is that the root r of the relator rn is a cyclically reduced word in A ∗ B, which is not a proper power. Hence without loss of generality, it has the form
r = a1b1 . . . a l b l , (4.2) 2 2 where {ai | 1 ≤ i ≤ l/2} = X ⊆ A, {bi | 1 ≤ i ≤ l/2} = Y ⊆ B and X ∪ Y does not contain the the identity element of A or B. Finally we assume that X generates A and Y generates B. This is a pretty standard n assumption, for suppose Freiheitssatz holds for Go = (Ao ∗ Bo)/N(r ) where Ao is
67 Chapter 4: One-relator product of two non-trivial groups with short relator
the subgroup of A generated by X and Bo is the subgroup of B generated by Y , then same is true for G = A ∗Ao Go ∗Bo B. Also if r has order n in Go, then it has order n in G.
4.2 Preliminary results
Most of the arguments we present involve analysis of zones incident on an interior vertex of M. We show that (l − 1)-zones do not occur, and if an (l − 2)-zone occurs, then there is only one possibility for its label. Later on l will either be 4, 6, or 8. Unless stated otherwise, we assume n ≥ 2. In what follows all subscripts are modulo l/2.
Lemma 4.2.1. Suppose an (l − 1)-zone Z is incident on some vertex, and none of the regions in Z is exceptional. Then one of the following holds.
1. r is a proper power.
2. n = 2 and there exist letters x, y ∈ A ∪ B with y2 = 1, and a W ∈ A ∗ B with `(W ) = l/2 − 1, such that r is conjugate to a word of the form xW yW −1.
Proof. Suppose that Z joins vertices of opposite signs. By Lemma 3.5.3, a cyclic subword of rn of length l + µ − 2 ≥ l + µ − gcd(l, µ) has period µ ≤ l/2. Hence by Theorem 3.2.5, r has period gcd(l, µ) since gcd(l, µ) ≥ 2. This implies that r is a proper power. On the other hand suppose that Z joins vertices of same sign. Up to cyclic relabelling of r, we may assume that the zone Z identifies b1a2b2 . . . al/2 with the subword −1 n −1 (aλbλ . . . bl/2+λ−2) of r for some 1 ≤ λ ≤ l/2 . In particular aj = al/2+λ−j for −1 2 ≤ j ≤ l/2 and bj = bl/2+λ−1−j for 1 ≤ j ≤ l/2 − 1. −1 −1 If λ > 2, then al/2+λ−1 = aλ−1 = al/2+λ+1−λ = a1. Similarly bl/2 = bλ−1. So we −1 can get an (l + 1)-zone by bridge-moves. If λ = 1, then al/2 = a1 . Similarly if −1 λ = 2, then b2 = bl/2−1. In either case we can do a bridge-move, changing Z to get an l-zone. Hence r has the form r = xW yW −1, with (x, y) admissible since y2 = 1. If n > 2, we can apply Theorems 1 and 3 in [62] to show that Freiheitssatz holds for G and r has order n in G. Since this contradicts our assumption, we must have that n = 2. This completes the proof.
Corollary 4.2.2. If an (l − 1)-zone Z whose regions are all interior joins vertices of same sign, then either the size of Z can be increased by bridge-moves or r has the form r = xW yW −1.
68 Chapter 4: One-relator product of two non-trivial groups with short relator
As a consequence, we have the following result which is a special case to a more gen- eral result proved in [[55],[56]]. However we include the proof here for completeness.
Theorem 4.2.3. Suppose n > 3 and `(r) ≤ 10, then Freiheitssatz holds for G. Also r has order n in G.
Proof. By Lemma 4.2.1, the number of edges incident on a vertex v is at most 5(l − 2) ≤ 4l ≤ nl. If any of the inequalities is strict, then there is nothing to prove. Hence we suppose that l = 10 and n = 4. In which case all the zones must have size l − 2. It easy to see that we can apply bridge-moves to show that some letter is trivial. This contradiction completes the proof.
Lemma 4.2.4. Let Z be an (l − 2)-zone joining two interior vertices. Suppose that Z can not be changed to (l − 1)-zone by a bridge-move. Then up to cyclic relabelling n of r, we may assume that the zone Z identifies the subword b1a2b2 . . . bl/2−1 of r ±1 ±n with the subword (bλaλ+1bλ+1 . . . bl/2+λ−2) of r , for some λ ≤ l/2, and one of the following holds:
1. If the two vertices have opposite signs, then B is cyclic. Furthermore neither of the two regions on both sides of Z is an m-gon for m < 4.
2. If the two vertices have same sign, then λ − 2 ≡ 1 mod l/2. In particular, the middle 2-zone of Z has both corners labelled by same letter.
Proof. First suppose that the two vertices have opposite signs. Then b1a2b2 . . . bl/2−1 is identified with the subword bλaλ+1bλ+1 . . . bl/2+λ−2. By Lemma 3.5.3, a cyclic n subword W of r of length `(W ) = l + µ − 3, with b1a2b2 . . . bl/2−1 as an initial or terminal segment, has period µ = 2(λ − 1) ≥ 2. Without loss of generality, we assume that W = b1a2b2 . . . bl/2+µ−2. If µ = 2, then B is cyclic. Also ai = ai+1 for 1 < i < l/2. If any of the two regions is an m-gon for n < 4, we get the relation ±1 a1al/2ai = 1, for some i. This will imply that A is cyclic, contradicting assumption. Suppose µ > 2. If gcd(l, µ) > 2 , then l+µ−3 > l+µ−gcd(l, µ). And so by Theorem 3.2.5, rn has period gcd(l, µ) ≤ µ/2. Hence r is a proper power. Otherwise suppose that gcd(l, µ) = 2. Then gcd(l/2, λ − 1) = gcd(l/2, µ/2) = gcd(l, µ)/2 = 1. But we have that bi = bi+λ−1 for 1 ≤ i ≤ l/2 − 1 and
l l l + λ − 2 = + λ − 1 − gcd , λ − 1 . (4.3) 2 2 2
So b1b2 . . . bl/2 has period 1. Hence B is cyclic.
Similarly, ai = ai+λ−1 for 2 ≤ i ≤ l/2 − 1. So we get l/2 − 2 equalities amongst l/2 elements. Any circuit will imply that m(λ − 1) ≡ 0 mod l/2 for some m < l/2.
69 Chapter 4: One-relator product of two non-trivial groups with short relator
This is not possible by assumption. Also either a1 and aλ or al/2 and al/2+λ−1 do not belong to same chain of equality. Hence without loss of generality a1aλai = 1 for some i. This implies A is cyclic.
Suppose the two vertices have same sign. In which case b1a2b2 . . . bl/2−1 is identified −1 with (bλaλ+1bλ+1 . . . bl/2+λ−2) , where λ ≤ l/2. It follows that for 2 ≤ i ≤ l/2 − 1 and 1 ≤ j ≤ l/2 − 1 −1 ai = al/2+λ−i (4.4) and −1 bj = al/2+λ−j−1 (4.5) respectively. There is nothing to prove when λ = 1. Also if λ = l/2, then by Equation 4.4 −1 a l = a1 . (4.6) 2 −1 Finally if 2 ≤ λ ≤ l/2 − 1, then −1 aλ = al/2. (4.7) In each case a bridge-move can be applied to increase the size of Z to l − 1. This contradicts hypothesis, hence the proof is complete.
One property we require of M is the existence of a positively-curved interior vertex, which we shall denote by ϑ. Throughout ϑ is assumed to have positive sign. This leads us to the following result.
Lemma 4.2.5. Let M be as above and let t be the minimum number of interior re- gions bounded by a boundary vertex which are not 2-gonal. Then one of the following holds
1. M has a positively curved interior vertex
2. If the exceptional region ∆o has degree k, then k(3 − t) ≥ 6. In particular k ≥ 4 and there exist a boundary vertex of degree 3.
Proof. Note that if M has an exceptional vertex vo (with label a proper subword of rn), then we can assume that all other vertices are interior. For each region ∆ of M of degree d(∆), assign an angle (d(∆) − 2)π/d(∆) to each of its d(∆) corners. Using the curvature formulas in Section 1.5.3 we have that regions of M have zero curvature. Since vo has curvature at most 2π, we must have a positively-curved interior vertex. Hence we assume that M has an exceptional region ∆o. We prove this Lemma by considering the degree d(∆o) of ∆o. By assumption, there are no interior regions ∆ with degree d(∆) = 1. Suppose that d(∆o) = 1. Then there is only one boundary vertex namely the vertex – v say –
70 Chapter 4: One-relator product of two non-trivial groups with short relator
bounding ∆o. All other vertices are interior. Since v has one region of degree 1 and (ln − 2) of degree at least 2, it follows that the curvature of v is at most 3π. But the total curvature of M is 4π, so there must be an interior vertex of positive curvature, as claimed.
Suppose then that d(∆o) = 2. Then there are no regions of degree 1, so any vertex v has curvature at most 2π, with equality only if every region bounded by v has degree 2. But if every region bounded by v has degree 2, then v has only one neighbouring vertex v0. Since M is W-minimal, it consists of only these two vertices. If this happens, then by the proof of Lemma 4.2.1, either r is a proper power or the label of ∆o was already trivial in A ∪ B. Either case we get a contradiction. Otherwise there are at most two vertices bounding ∆o, each with curvature strictly less than 2π. Since the total curvature of M is 4π, there must be a positively curved interior vertex, as claimed.
Suppose then that ∆o has degree d(∆o) = k ≥ 3. Then there are no regions of degree 1. For each vertex v bounding ∆o, let τ(v) be number of regions it bounds outside ∆o which are 2-gonal. It follows that any such vertex has curvature
τ(v) (k − 2) κ(v) ≤ 2 − − π. (4.8) 3 k If there are no positively-curved interior vertices, then
X t (k − 2) 4π ≤ κ(v) ≤ 2 − − kπ, (4.9) 3 k where the summation is over all boundary vertices. It follows that 6 ≤ k(3 − t). This can only happen if t ≤ 2 and k ≥ 3. Note that none of the zones occurring can have size greater than l by minimality. Hence in particular that t 6= 0. If k = 3, then there are no interior vertices and all three boundary vertices bound exactly two regions which are not 2-gonal. In other words M consists of the three boundary vertices and the l-zones joining them as shown in Figure 4.1.
Figure 4.1: A picture consisting of only k boundary vertices joined to each other by l-zones.
71 Chapter 4: One-relator product of two non-trivial groups with short relator
3 However each of the two 3-gons (one of which is ∆o) has label x . Since one is 3 interior, it follows that x = 1. In other words the label of ∆o was already trivial in G. Hence k ≥ 4 and there exist a boundary vertex of degree 3. This completes the proof.
Remark 4.2.6. It is easy to see that the existence of a boundary vertex of degree 3 in Lemma 4.2.5 implies the existence of a positively-curved boundary vertex of degree 3.
Corollary 4.2.7. Suppose an (l − 3)-zone Z joins ϑ to a vertex u of the same sign, with l ≥ 6. Then
1. either the size of Z can be increased by bridge-moves, or
−1 −1 2. l = 6 and without loss of generality, r = a1b1a2b2a2 b1 .
Proof. As in Lemma 4.2.4, without loss of generality Z identifies b1a2b2 . . . bl/2−2al/2−1 with aλbλ . . . al/2+λ−3bl/2+λ−3, where λ ≤ l/2. Hence for 2 ≤ i ≤ l/2 − 1 and 1 ≤ j ≤ l/2 − 2, −1 ai = al/2+λ−i−1 (4.10) and −1 bj = bl/2+λ−2−j (4.11)
−1 respectively. By Equation 4.10, al/2−1 = a1 when λ = 1. Similarly by Equation −1 4.11, b1 = bl/2−1 when λ = 2. In either case a bridge-move can be done on the left side or right side of Z, increasing its size to l − 2. When l = 6, the only other −1 possibility is λ = 3. In this case Equations 4.10 and 4.11 reduces to ai = a5−i and −1 −1 bj = b4−j, where 2 ≤ i ≤ 3 and 1 ≤ j ≤ 2 respectively. In other words, a2 = a3 , −1 −1 −1 −1 b1 = b3 and b2 = b2 . Hence r = a1b1a2b2a2 b1 .
−1 Now suppose that l > 6 and λ > 2, then bλ−1 = bl/2−1 by Equation 4.11. As before, we can do a bridge-move to increase the size of Z. This completes the proof.
We also include the following trivial results which shall be used later.
Proposition 4.2.8. The group H = hx, y | x2 = 1, w(x, y) = 1i with w(x, y) ∈ {xy±1xy2, xyxy} is either dihedral or cyclic.
This can be proved by applying Tietze transformations. In fact H is the cyclic group of order 6 when w(x, y) = xyxy2, dihedral group of order 6 when w(x, y) = xy−1xy2 and infinite dihedral when w(x, y) = xyxy. We are also interested in the case where w(x, y) = xyxy−1. In other words, H is abelian.
72 Chapter 4: One-relator product of two non-trivial groups with short relator
2 Lemma 4.2.9. Let G = (A ∗ B)/N(r ) where A is Abelian with factor groups Ai, S and B has a faithful representation in PSL2(C). If pi(r) is neither in i N(Ai) nor
N(B), where pi : A ∗ B → Ai is the canonical projection map, then G satisfies the Freihetssatz.
Proof. For each i, both Ai and B have faithful representations in PSL2(C). Hence both embed in G since they embed in pi(G). Moreover
\ Ker(pi) ∩ A = 1, so A also embeds in G.
We now focus on positively-curved interior vertices ϑ. All the zones mentioned here are incident on ϑ. Recall that a vertex of degree m has m zones incident on it. Let the zones be Z1,. . . , Zm listed consecutively in clockwise order.
Definition 4.2.10. A configuration of a vertex of degree m is the m-tuple (ω1, . . . , ωm) or any of its cyclic permutations, where ωi is the size of Zi.
A bridge-move induces a map from one configuration of a vertex to another. Since we will be adopting a rigorous method which involves checking all configurations, we can try to that do that effectively. One way is to consider two configurations equal up to orientation and cyclic permutation. For example (6, 5, 5, 4, 4), (6, 4, 4, 5, 5) and (4, 6, 5, 5, 4) are all equal.
4.3 Relator of length eight with power n ≥ 3
In the section we consider the one-relator group
(A ∗ B) G = , (4.12) N(rn) where n ≥ 3 and r = axbyczdw with a, b, c, d ∈ A and x, y, z, w ∈ B. The aim is to prove the following theorem.
Theorem 4.3.1. Let G be as in 4.12. Then the following holds for G.
1. G satisfies the Freiheitssatz.
2. The word r has order n in G.
To prove Theorem 4.3.1, it is enough by Theorem 4.2.3 to restrict to the case of n = 3. Suppose by contradiction that the Theorem fails. Then we obtain a non-trivial W-minimal spherical picture M over G (with exceptional vertex or region as the case
73 Chapter 4: One-relator product of two non-trivial groups with short relator
Type 1 Type 2 Type 3 Type 4 (6, 6, 6, 6) (6, 3, 6, 5, 4) (6, 5, 5, 5, 3) (6, 5, 4, 4, 5) (6, 6, 1, 6, 5) (6, 4, 6, 3, 5) (6, 5, 5, 4, 4) (6, 6, 2, 6, 4) (6, 4, 6, 4, 4) (6, 5, 5, 3, 5) (6, 6, 3, 6, 3) (6, 5, 6, 3, 4) (6, 4, 5, 5, 4) (6, 6, 3, 5, 4) (6, 5, 6, 2, 5) (6, 5, 5, 6, 2) (6, 6, 4, 4, 4) (6, 4, 5, 4, 5) (6, 6, 5, 2, 5) (6, 6, 5, 3, 4) (6, 6, 5, 4, 3) (6, 6, 5, 5, 2)
Table 4.1: A table containing all the possible configurations of a positively-curved interior vertex. may be). To each region ∆ in M of d(∆) we assign an angle of (d(∆) − 2)π/d(∆). By Lemma 4.2.5, we deduce that M contains a positively-curved interior vertex ϑ i.e. d(ϑ) ≤ 5. If d(ϑ) ≤ 3, then the result follows from Lemma 4.2.1. Hence d(ϑ) = 4 or 5. We list all the possible configurations of ϑ (see Table 4.1) and show that each of them leads to a contradiction to the general assumptions. More precisely, analysis of these configurations will show that each of A and B has faithful representation in PSL2(C). Throughout this Section we shall use the letter t (with or without subscript) to refer to an element in {a, b, c, d}±1, and s (with or without subscript) to refer to an element in {x, y, z, w}±1. With the exception of (6, 6, 6, 6), all other configurations in Table 4.1 has degree five. Since ϑ has positive curvature, it follows that at least four of the five regions bounded by ϑ which are not 2-gons must be 3-gons. Hence by Lemma 4.2.4, 6-zones of a configuration of degree five joins vertices of same sign. Without loss of generality we assume that at least one the 6-zones in any configuration has corners at ϑ labelled x, b, y, c, z respectively in the clock-wise direction. So again by Lemma 4.2.4, we get the relations xz = bc = y2 = 1. Note that the same assumptions hold even in the case of (6, 6, 6, 6) by restricting to the 6-zones bounding a 3-gon.
Lemma 4.3.2. Configurations of Type 1 do not occur.
Proof. The proof uses only the fact in each of the configurations in Type 1, there are two adjacent 6-zones and at least one of A-regions bounded by these zones is a 3-gon. In the case of (6, 6, 6, 6), we choose the two 6-zones bounding a 3-gon. By Lemma 4.2.4 these 6-zones connect vertices of same sign. So without loss of
74 Chapter 4: One-relator product of two non-trivial groups with short relator generality we can assume that the two adjacent 6-zones are labelled as shown in Figure 4.2 below.
Figure 4.2: Diagram showing adjacent 6-zones of a configuration of Type 1
2 2 So in B the relations xz = y = yw = x = 1 hold. Hence B is a quotient of D∞. In A we have the relations bc = ab = 1 and tcd = 1 with t ∈ {b±1, a, c, d}, since one of the regions with a corner labelled c must be a 3-gon. Hence A is cyclic, contradicting assumptions.
Lemma 4.3.3. Configurations of Type 2 do not occur.
Proof. Every configuration in Type 2 has the form (6, 3, 6, ∗, ∗), (6, 4, 6, ∗, ∗), or (6, 5, 6, ∗, ∗), where ∗ is a place holder. We shall treat each of the three cases separately beginning with (6, 3, 6, ∗, ∗).
1. Without loss of generality Figure 4.3 gives a labelling of ϑ in case of (6, 3, 6, ∗, ∗).
Figure 4.3: Diagram showing configuration (6, 3, 6, ∗, ∗) in Type 1
We can assume that A-region and the B-region containing t and s respectively are 3-gons. In other words tad = 1 = sxw. The two 6-zones give the relations xz = y2 = yz = 1 and bc = bd = c2 = 1. Hence both A and B are cyclic which contradicts assumption.
2. In the case of (6, 4, 6, ∗, ∗) we can assume without loss of generality that ϑ is labelled as shown in Figure 4.4.
75 Chapter 4: One-relator product of two non-trivial groups with short relator
Figure 4.4: Diagram showing configuration (6, 4, 6, ∗, ∗) in Type 2
The two 6-zones give the relations bc = cd = 1 and xz = y2 = yw = z2 = 1.
Hence we conclude that B is a quotient of D∞. Also we can assume that tad = 1. It follows that A is cyclic. This contradicts assumption.
3. Finally in the case of configuration (6, 5, 6, ∗, ∗) we can assume without loss of generality that ϑ is labelled as shown in Figure 4.5.
Figure 4.5: Diagram showing configuration (6, 5, 6, ∗, ∗) in Type 2
Hence the two 6-zones give the relations ac = bc = d2 = 1 and xz = zw = y2 = 1. Also we can assume as before that tad = 1 = sxy. Hence both A and B are cyclic, a contradiction to assumption.
Lemma 4.3.4. Configurations of Type 3 do not occur.
Proof. Configurations in Type 3 can be grouped essentially in two: configurations of the form (6, 4, 5, ∗, ∗) or (6, 5, 5, ∗, ∗). We assume that the middle 5-zone can not be changed to a 6-zone via a bridge-move, as this will take us back to Type 2 for which the result already holds. Hence the middle 5-zones joins vertices of opposite signs by Lemma 4.2.7. We begin with case (6, 4, 5, ∗, ∗).
1. Without loss of generality Figure 4.6 shows all the possible labellings of 6-zone and the middle 5-zone of configuration (6, 4, 5, ∗, ∗).
76 Chapter 4: One-relator product of two non-trivial groups with short relator
Figure 4.6: Diagram showing configuration (6, 4, 5, ∗, ∗) in Type 3
The 6-zone gives the relations xz = y2 = 1 and bc = 1. The 5-zone gives three possible set of relations.
(a) In the first case x = y = z and b = c = d. Since at = 1, A is either cyclic
or a quotient of D∞. Similar conclusion holds for B since ws = 1. (b) In the second case y = z = w and c = d = a. Hence both A and B are cyclic. (c) In the third case y = w, x = z and a = c, b = d. Hence B is a quotient
of D∞ and A is cyclic.
In each case we get a contradiction to assumption.
2. Figure 4.7 shows all the possible labellings of 6-zone and the middle 5-zone of configuration (6, 5, 5, ∗, ∗).
Figure 4.7: Diagram showing configuration (6, 5, 5, ∗, ∗) in Type 3
The 6-zone gives the relations xz = bc = y2 = 1. The 5-zone gives three possible set of relations.
(a) In the first case y = z = w and b = c = d. So B is cyclic and A is either
cyclic or a quotient of D∞ (since at = 1). (b) In the second case x = z = w and c = d = a. Hence A is cyclic and B
quotient of D∞.
77 Chapter 4: One-relator product of two non-trivial groups with short relator
(c) In the third case y = w, x = z and a = c, b = d. Hence B is a quotient
of D∞ and A is cyclic.
In each case we get a contradiction to assumption.
Lemma 4.3.5. Configurations of Type 4 do not occur.
Proof. There is only one configuration of Type 4 namely (6, 5, 4, 4, 5). We assume that no bridge-moves can be done to change any of the two 5-zones to a 6-zone. For otherwise we get one of configurations (6, 5, 5, 4, 4) or (6, 6, 3, 4, 5) for which the result is already known. As before both 5-zones connect vertices of opposite signs. In Figure 4.8, we have shown all the possible labellings of ϑ with this configuration, where the 2-zone indicated is part of the second 5-zone.
Figure 4.8: Diagram showing configuration (6, 5, 4, 4, 5) in Type 4
The 6-zone gives the relations xz = bc = y2 = 1. The 5-zone gives three possible set of relations. Also td = 1 where t ∈ {a−1, b−1, c−1}.
1. In the first case x = y = w and a = b = c. So both A and B are cyclic.
2. In the second case y = w, x = z and a = c, b = d. Hence B is a quotient of
D∞ and A is cyclic.
3. In the third case x = z = w and a = b = d. Hence A is cyclic and B quotient
of D∞.
In each case we get a contradiction to assumption that at least one of A or B does not have a faithful representation in PSL2(C).
This shows that our assumption that Theorem 4.3.1 fails was false, hence the proof of the Theorem 4.3.1 is complete.
78 Chapter 4: One-relator product of two non-trivial groups with short relator
4.4 Relator of length six with power n ≥ 2
In the section we consider the one-relator group
(A ∗ B) G = , (4.13) N(rn) where n ≥ 2 and r = axbycz with a, b, c ∈ A and x, y, z ∈ B. As in the case of l = 8, we list all the possible configurations of a positively-curved interior vertex. However things are a little more complicated in this case. We call a configuration of a positively-curved interior vertex good if it is not of the form (4, 2, 2, 2, 2), (4, 2, 4, 1, 1) or (4, 2, 4, 2), and not good (or bad) otherwise. We say that a picture M is not good (or bad) if the configuration of each positively-curved interior vertex of M is not good. The main result of this section is the following.
Theorem 4.4.1. Let G be as in 4.13. Then either:
1. A and B have faithful representations in PSL2(C), or
2. n = 2 and up to conjugation r has the form r = axbx−1cz with z2 = 1, where a, b, c ∈ A and x, z ∈ B.
We can restrict to the case where n = 2 for the following reasons. Suppose n > 2 and M does not have a positively-curved interior vertex ϑ. Then there is a boundary vertex v of degree 3 by Lemma 4.2.5. Hence v bounds an ω-zone with ω ≥ 5. Same conclusion holds if M has a positively-curved interior vertex ϑ of degree less than 5 or n > 3. In these cases the result follows from Lemma 4.2.1 and Lemma 4.4.2 below. Hence we suppose that n = 3 and ϑ has degree 5 and a configuration of the form (4, 4, ∗, ∗, ∗). In other words, there are two adjacent 4-zones incident on ϑ. Without loss of generality and using Lemma 4.2.4, we assume the two adjacent 4-zones are labelled as shown in Figure 4.9.
Figure 4.9: Configuration (4, 4, ∗, ∗, ∗)
We deduce from the two 4-zones that xy = xz = 1 and a2 = b2 = 1. In particular B is cyclic. Also at least one of bac = 1 or tac = 1, with t ∈ {a, c} holds. So A is cyclic or a quotient of D∞. Hence the Theorem holds. For the rest of this section we assume that n = 2.
79 Chapter 4: One-relator product of two non-trivial groups with short relator
We assume that Theorem 4.4.1 fails and proceed to derive a contradiction. We do this in a series of Lemmas. In particular, most of our conclusions will follow from Lemma 4.4.2 below.
Lemma 4.4.2. If r has the form r = axbyb−1x−1, then G = (A ∗ B)/N(rn) satisfies the Freiheitssatz and r has order n.
Proof. Suppose r = axbyb−1x−1 and either G fails to satisfy the Freiheitssatz or r does not have order n in G. We construct a non-trivial W-minimal spherical clique-picture M over G. A typical clique-label will have the form
aα1 (xb)yβ1 (b−1x−1) . . . aαk (xb)yβk (b−1x−1).
So G is induced by the generalised triangle group
T (p, q, 2) = ha, y | ap, yq, (ay)ni, where p and q are the orders of a and y respectively. Hence by the Spelling Theorem for generalised triangle groups, it follows that k ≥ n. In any region of M, we may assume that no two consecutive corner labels are of the form aα or yβ (for otherwise we could combine the two cliques into one, forming a smaller clique-picture). In particular, if a 2-zone is labelled by aα at one corner, the opposite corner has to be a b±1 and so A will be cyclic. Same argument applies to B. The results holds if both A and B are cyclic. Hence we can assume without loss of generality that A is not cyclic and proceed to derive a contradiction. There are at least two corners labelled by aα and atleast two corners labelled yβ (since n ≥ 2). This implies by Lemma 4.2.5 that M has a positively-curved interior clique ϑ. Curvature argument tells us that in any positively-curved interior clique, there are at most five regions which are not 2-gons. Since A is not cyclic, no A-region with corner labelled aα is a 2-zone. Suppose a 2-zone has a corner labelled b (respectively b−1), then the opposite corner has label b (respectively b−1). If not, we get a contradiction to minimality or the fact that A is not cyclic. Consequently, there is no zone with label (byβi b−1)±1. Hence the two A-regions with corner labelled by aα and at most one in the sequence x, b, yβk , b−1, x−1 are contained in regions which are not 2-gons. This implies (wlog) that there is a 3-zone with corners at ϑ labelled x and b respectively. Hence b2 = 1 in A, and B is cyclic since x must be a power of y. Also there are at least four regions which are not 2-gonal and any A-region with corner labelled aα is at least a 4-gon. Since ϑ is assumed to be positively-curved, it must have degree four. Hence at least one of these regions which are not 2-gons must be a 3-gon. Since no zone
80 Chapter 4: One-relator product of two non-trivial groups with short relator
has label (byβi b−1)±1, any such 3-gon must have two consecutive corners labelled x, x−1 or b±1, aαi . Both cases gives a contradiction. Hence the proof.
Using Lemma 4.2.1 and Lemma 4.4.2, we can assume that there are no ω-zone joining two vertices of same sign with ω > 4, and all the 2-zones are interior. Hence we can easily deduce that M contains a positively-curved interior vertex. As before we shall use the letter t (with or without subscripts) to refer to an element in {a, b, c}±1, and s (with or without subscripts) to refer to an element {x, y, z}±1.
Lemma 4.4.3. M has a positively-curved interior vertex ϑ.
Proof. Suppose by contradiction that M has no positively-curved interior vertex. Since every vertex has degree at least 3, it follows from the proof of Lemma 4.2.5 that the exceptional region has degree at least 6. Also we have that M contains a positively-curved boundary vertex v of degree 3 which bounds at least one interior region of degree at most 4 (for otherwise curvature of M is less than 4π). By the comments succeeding Lemma 4.4.2, we assume that all the zones have equal size which is 4, and that no bridge-moves can be applied to increase the size of any of the zones. We can arrange so that the middle 2-zones have corners labelled a, c and b consecutively in the clockwise-sense as shown in Figure 4.10. Hence the three regions which are not 2-gonal are all A-regions.
Figure 4.10: Diagram showing positively-curved boundary vertex of degree 3
If all three 4-zones connect vertices of opposite signs, then by Lemma 4.2.4, xy = xz and a2 = b2 = c2 = 1. In particular B is cyclic. Also we have a relation of length at most 5 involving a, b, and c. So A is a quotient of D∞. Hence we suppose that at least one of the 4-zones joins vertices of opposite signs. It follows from Lemma 4.2.4 that B is cyclic. If exactly one 4-zone joins to a vertex of negative sign, then without loss of generality a2 = b2 = 1 and t ∈ {a−1, b−1}. Hence A is dihedral. If exactly two 4-zones joins to
81 Chapter 4: One-relator product of two non-trivial groups with short relator a vertex of negative sign, then without loss of generality we assume that a2 = 1 and b = a or b = c. Similarly c = a or c = b. Any of the possibilities must imply that A 2 −1 is cyclic, except when a = bc = 1 are the only relations. In other words t1 = b, −1 −1 −1 −1 t2 = a, t3 = a , t4 = c , t5 = b and t6 = a . If any of the three regions is interior and has degree 3, then A is cyclic. So we conclude that there is a relation of the form ab2p = 1, babp = 1 or bapa = 1, for some p ∈ {a, b, c}±1. Hence A is either cyclic, dihedral or abelian. So G satisfies the Freiheitssatz. Clearly r has order 2 in
G since B is isomorphic to Z2. Finally, if all 4-zones connect to vertices of negative signs, then it is easy to see that a = b = c, and so A is cyclic.
Type 1 Type 2 Type 3 Type 4 Type 5 Type 6 (4, 3, 4, 1) (3, 4, 2, 1, 2) (4, 4, 2, 1, 1) (2, 3, 2, 4, 1) (3, 3, 2, 2, 2) (4, 2, 2, 2, 2) (3, 3, 3, 3) (3, 4, 2, 2, 1) (4, 4, 1, 2, 1) (2, 2, 3, 1, 4) (3, 2, 3, 2, 2) (4, 2, 4, 1, 1) (4, 3, 3, 2) (3, 4, 1, 3, 1) (4, 1, 4, 2, 1) (2, 4, 2, 1, 3) (3, 3, 3, 2, 1) (4, 2, 4, 2) (4, 3, 2, 3) (3, 4, 3, 1, 1) (1, 4, 1, 3, 3) (3, 3, 1, 3, 2) (4, 4, 2, 2) (3, 4, 1, 1, 3) (4, 4, 3, 1) (3, 4, 1, 2, 2)
Table 4.2: A table containing all the possible configurations of a positively-curved interior vertex.
Table 4.2 contains all the possible configurations of ϑ. We assume that at least one of the 4-zones in each configuration has corners at ϑ labelled x, b and y consecutively. So such a 4-zone gives the relations xy = b2 = 1 (if it joins vertices of same sign) or B is cyclic and b = a or c (if it joins vertices of opposite signs) by Lemma 4.2.4. The proof on Theorem 4.4.1 is divided in two parts depending on whether or not M is good.
Part 1
Here we consider the case where M is good and we prove that part 1 of Theorem 4.4.1 holds.
4.4.1 Configurations of an interior vertex of degree 4
In this section we prove that the existence of a positively-curved interior vertex of degree 4 implies Theorem 4.4.1. We do this in a series of Lemmas, each dealing
82 Chapter 4: One-relator product of two non-trivial groups with short relator with a specific configuration in column 1 of Table 4.2. We begin with configuration (4, 4, 3, 1).
Lemma 4.4.4. Theorem 4.4.1 holds in the case of configuration (4, 4, 3, 1)
Proof. First suppose that each of the two 4-zones joins vertices of same sign. Then from the two 4-zones we get the relations xy = xz = 1 and a2 = b2 = 1. So in particular B is cyclic. We assume that c 6∈ ha, bi, as otherwise A is a quotient of
D∞. In which case the result holds. It follows that the 3-zone must be labelled as shown in Figure 4.11.
Figure 4.11: Configuration (4, 4, 3, 1)
Hence the B-region can not be a 3-gon, so at least one of the A-regions is a 3-gon. Hence we get a relation of length at most 3 involving c and at least one of a or b. It follows that A is cyclic or dihedral. Suppose that one of the 4-zones joins vertices of opposite signs. Then without loss of generality B is cyclic and b ∈ ha, ci. Since the other 4-zone and the 3-zone have corners labelled a and c respectively, A is either cyclic or a quotient of D∞. This is a contradiction, hence the proof.
Lemma 4.4.5. Theorem 4.4.1 holds in the case of configuration (4, 3, 4, 1)
Proof. Suppose that each of the two 4-zones joins vertices of same sign as shown in Figure 4.12. Then from the two 4-zones we get the relations xy = y2 = 1 and bc = b2 = 1. If the 3-zone joins vertices of same sign, then one of xz = 1, ab = 1 = yz, or ac = 1 holds. Also if exactly one of the two 4-zones joins vertices of same sign, then by symmetry we can assume that x = y = z, b = a or b = c, and bc = y2 = 1. In each case a cyclic conjugate of r satisfies the conditions of Lemma 4.4.2. The result follows.
83 Chapter 4: One-relator product of two non-trivial groups with short relator
Figure 4.12: Configuration (4, 3, 4, 1)
If each of the two 4-zones joins vertices of opposite signs, then both A and B are cyclic. The result follows by contradiction.
Lemma 4.4.6. Theorem 4.4.1 holds in the case of configuration (4, 4, 2, 2)
Proof. Suppose that each of the two 4-zones joins vertices of same sign. Then xy = xz = 1 and a2 = b2 = 1. Hence in particular B is cyclic.
Figure 4.13: Configuration (4, 4, 2, 2)
If c ∈ ha, bi, then we are done. Otherwise we can assume that the only triangular 3 region is the one with corner labelled t6 with t6 = t7 = c. Hence c = 1. Without loss of generality we assume that t1tt8a = ctba = 1 for some t ∈ {a, b, c}. For each possibility A is a quotient of D∞. Suppose that each of the two 4-zones joins vertices of opposite signs. Then B is −1 −1 −1 −1 cyclic as well as A, unless t1 = c , t2 = b , t3 = a and t4 = c . As before we can assume that the only triangular region is the one with corner labelled t6 with 3 t6 = t7 = c. Hence c = 1. Without loss of generality we assume that t1tba = 1, where t = c−1 or a (alternatively b). Either possibility implies A is cyclic. Finally, we assume that only one of the two 4-zones joins vertices of opposite signs. −1 −1 Hence without loss of generality t1 = c , t2 = b , t3 = b and t4 = c. Hence
84 Chapter 4: One-relator product of two non-trivial groups with short relator
−1 2 ab = a = 1 and B = Z2. It follows that a cyclic conjugate of r has the form x(by)c(za) = x(by)c(y−1b−1). Hence the result follows from Lemma 4.4.2.
Lemma 4.4.7. Theorem 4.4.1 holds in the case of configuration (4, 3, 3, 2)
Proof. Suppose the 4-zone joins vertices of same sign, then t4 = a, t3 = c and xy = b2 = 1. By Lemma 4.4.2 we can assume that ac 6= 1.
Figure 4.14: Configuration (4, 3, 3, 2)
1. If the 3-zone adjacent to the 4-zone joins vertices of same sign, then t5 =
c, t6 = x and zy = ab = 1. Note that if t5 6= a by assumption and if t5 = b, then a2 = xz = 1 and so in particular B is cyclic. Also we can assume that the
only triangular region is the one with corner labelled t2. So t2ac = 1 = tt4ct5. It follows that A is cyclic.
Hence the result follows from Lemma 4.4.2.
2. Suppose then that the 3-zone adjacent to the 4-zone joins vertices of opposite signs. There are two cases to consider.
−1 −1 −1 −1 (a) If t5 = b , t6 = z and zy = ac = 1. Then in particular B is cyclic. So we assume that b 6∈ ha, ci. This implies that the other 3-zone joins
vertices of same sign with t7 = y, t8 = a and so no new relations. However in such case the B-region can not be a 3-gon (as otherwise z = 1). Hence we get a relation act = 1, and so A is cyclic.
−1 −1 −1 −1 (b) If t5 = a , t6 = y and zx = ab = 1. Then the result follows from Lemma 4.4.2.
Suppose the 4-zone joins vertices of opposite signs, then B is cyclic and we can assume that none of the 3-zones joins vertices of same sign as that will imply that
85 Chapter 4: One-relator product of two non-trivial groups with short relator
−1 2 −1 2 either A is also cyclic or B = Z2 and ab = b = 1 (alternatively cb = b = 1). The result follows from Lemma 4.4.2.
−1 −1 1. If t3 = b , t4 = a and hence b = c, then we conclude that a ∈ hb, ci using the 3-zone with corner labelled a.
−1 −1 −1 −1 2. If t3 = c , t4 = b and hence b = a, we can assume that t5 = a , t6 = y , −1 −1 t7 = z , t8 = b . It follows that the B-region can not be a 3-gon (since x = y = z). Hence one of the A-regions is a 3-gon and since each has corners labelled c±1 and a±1 or b±1, A is cyclic.
The result follows by contradiction.
Lemma 4.4.8. Theorem 4.4.1 holds in the case of configuration (4, 3, 2, 3)
Proof. Suppose the 4-zone joins vertices of same sign, then xy = b2 = 1.
Figure 4.15: Configuration (4, 3, 2, 3)
Suppose the 3-zone on the right joins vertices of same sign. Then we can apply Lemma 4.4.2 except in the case where a2 = xz = 1. Similarly we can apply Lemma 4.4.2 if the 3-zone on the right joins vertices of same sign except when c2 = yz = 1.
So if both 3-zones connect vertices of same sign, then B = Z2. Hence abc = 1 and so A is a quotient of D∞. Same conclusion holds if just one of the 3-zones joins vertices of opposite signs. Hence we assume that both 3-zones connect vertices of opposite signs. If the two 3-zones both give the relation ac−1 = xz−1 = yz−1 = 1 as shown in Figure 4.15, then B = Z2. Hence neither of the two B-regions can be triangular. So A is cyclic as abc = 1. Otherwise B is still cyclic and A is cyclic or a quotient of D∞. Finally, if the 4-zone joins vertices of opposite signs. Then by symmetry we can assume from the 4-zone that x = y = z and a = b. Since the two 3-zones have corners labelled a and c, A is either cyclic or a quotient of D∞ unless the two 3- 2 −1 −1 zones give the relations yz = c = 1 and xz = ab = 1. But in this case B = Z2
86 Chapter 4: One-relator product of two non-trivial groups with short relator and so for similar reason as before, we conclude that capa = 1, where p = 1 or ±1 p ∈ {a, b, c} . Hence A is cyclic or a quotient of D∞.
Lemma 4.4.9. Theorem 4.4.1 holds in the case of configuration (3, 3, 3, 3)
Proof. Suppose all the zones connect vertices of opposite signs.
Figure 4.16: Configuration (3, 3, 3, 3)
−1 −1 If t1 = c , t2 = x , then x = z and b = a. We can conclude that A is cyclic (since −1 −1 c is a corner label of some 3-zone). If t3 = x , then z = y and so B is cyclic. −1 −1 Hence we suppose that t3 = z and so t4 = b . It follows that none of the two A-regions can be a 3-gon, and each B-region has a corner labelled y and x−1 or z−1. Hence B is cyclic. The second case is similar. Hence in this case the result follows by contradiction. So we are reduced to the case where at least one of the zones joins vertices of same sign. By symmetry we can take this zone to be the ‘northern’ 3-zone. Suppose A is neither cyclic nor dihedral. Then by Lemma 4.4.2, we can assume that ac = 1 and xy = 1 (alternatively ab = 1 and yz = 1) do not hold simultaneously. In particular this implies that either t1 = b or t1 = c.
If t1 = b, then t2 = t3 = t7 = z. It follows that none of the four regions bounded by ϑ can be a 3-gon. This contradicts the assumption that ϑ is positively-curved.
Hence t1 = c. It follows that either
1. t2 = t6 = x, t3 = t7 = z, t4 = t8 = b, and t5 = c, or
−1 −1 2. t2 = t6 = x, t3 = t7 = z , t4 = t8 = b , and t5 = c.
In both cases the two B-regions can not be 3-gons and so at least one of the two A-regions is a 3-gon. So bac = 1. This contradicts the assumption that A is neither cyclic nor dihedral. By symmetry, B is also cyclic or dihedral. This completes the proof.
87 Chapter 4: One-relator product of two non-trivial groups with short relator
4.4.2 Configurations of an interior vertex of degree 5
In this section we consider positively-curved interior vertices ϑ of degree 5. It follows from Corollary 4.4.2 that 4-zones connect vertices of same sign.
Lemma 4.4.10. Configurations of Type 2 do not occur.
Proof. In each case, a 4-zone is adjacent to a 3-zone. The 4-zone joins vertices of same sign and is labelled as shown in Figure 4.17. Hence we get the relations xy = b2 = 1. So we assume by Lemma 4.4.2 that ac 6= 1.
Figure 4.17: Configuration of Type 2
−1 1. When t1 = a or t1 = c, we have ab = yz = 1. Hence the result follows from Lemma 4.4.2.
−1 2. When t1 = b or t1 = b , the B-region with corner labelled t2 can not be a 3-gon. Hence in the first case a2 = xz = acb = 1. In the second we have −1 −1 −1 yz = ac = acb = 1. So A is either cyclic or a quotient of D∞.
Lemma 4.4.11. Configurations of Type 3 do not occur.
Proof. The proof is divided into three cases: the case with adjacent 4-zones, the case with non-adjacent 4-zones, and the case with one 4-zone.
1. Suppose that ϑ has two adjacent 4-zones (as in (4, 4, 2, 1, 1) and (4, 4, 1, 2, 1)). Then from Figure 4.18, B is cyclic and b2 = a2 = 1.
Figure 4.18: Adjacent 4-zones
88 Chapter 4: One-relator product of two non-trivial groups with short relator
Also tca = 1 where t ∈ {a±1, b±1, c±1}. Hence A is cyclic or dihedral.
2. Suppose that ϑ has configuration (4, 1, 4, 2, 1). Then from Figure 4.19, the 4-zones give the relations ab = b2 = 1 and x2 = xy = 1.
Figure 4.19: Two 4-zones separated by a 1-zone
Also we have relations szx±1 = 1 and tac = 1. So both A and B are cyclic, hence the result.
3. Finally, suppose that ϑ has configuration (4, 1, 3, 3, 1). Then from Figure 4.20, the 4-zones give the relations xy = b2 = 1.
Figure 4.20: Configuration (4, 1, 3, 3, 1)
2 (a) When t1 = x, we have the relations a = xz = 1 and tac = 1. Hence both A and B are cyclic.
(b) When t1 = y, then the A-region containing t2 can not be a 3-gon. Hence szy = 1 and tac = 1. So each of A and B is either cyclic or dihedral. −1 −1 −1 Similar conclusion holds when t1 = x (since ab = xy = 1).
(c) When t1 = z, then we have ac = xy = 1. The result for this case follows from Lemma 4.4.2.
89 Chapter 4: One-relator product of two non-trivial groups with short relator
−1 −1 −1 (d) When t1 = y , we have ac = xz = 1, and so both A and B are cyclic.
Lemma 4.4.12. Configurations of Type 4 do not occur.
We prove this Lemma in a series of propositions one for each configuration in Type 4.
Proposition 4.4.13. Configuration (4, 1, 2, 3, 2) does not occur.
Proof. Figure 4.21 shows the labelling of ϑ with configuration (4, 1, 2, 3, 2). From the 4-zone we get the relation xy = b2 = 1. Also at = 1 = sz and at least one of t1ac = 1 or t2ac = 1 holds.
Figure 4.21: Configuration (4, 1, 2, 3, 2)
We consider the 3-zone.
−1 −1 −1 1. When s4 = z , then ab = xy = 1. So A is cyclic and B is either cyclic or dihedral.
−1 −1 −1 2. When s4 = y , we have bc = yz = 1. Hence B is cyclic and A is cyclic 2 or dihedral. Similar conclusion holds when s4 = z (since y = bc = 1).
3. When s4 = x, we have yz = bc = 1. In this case the result follows from Lemma 4.4.2.
4. When s4 = y, we get no new relations from the 3-zone. However we conclude ±1 that the B-region containing s4 is not a 3-gon. So if s 6= z, then t1 = b or ±1 t2 = b , and hence B is cyclic and A is either cyclic or dihedral. Hence we 2 suppose that s = z, t1 = c, t2 = a, so in particular z = 1 and a = c. We ±1 conclude from zs2s1 = 1 that B is cyclic (since s1, s2 ∈ {x, y} ). Also we can
90 Chapter 4: One-relator product of two non-trivial groups with short relator
−1 −1 assume that t = c , s2 = y and so s1 = x and t2 = b. Hence A is also cyclic
(since t2ac = 1).
Proposition 4.4.14. Configuration (4, 1, 3, 2, 2) does not occur.
Proof. Figure 4.22 shows the labelling of ϑ with configuration (4, 1, 3, 2, 2). From the 4-zone we get the relations xy = b2 = 1. Also bt = 1 = sz, and at least one of t1ac = 1 or t2ac = 1 holds. As before we assume ac 6= 1 by Lemma 4.4.2, so in particular s2 6= z.
Figure 4.22: Configuration (4, 1, 3, 2, 2)
We consider the 3-zone.
2 1. When s2 = x and so t3 = c, then xz = a = 1, and so in particular B is cyclic.
It follows that the B-region containing s2 is not a 3-gon. So A is cyclic (since
bt4t3 = 1).
−1 −1 −1 −1 2. When s2 = x and so t3 = c , then ab = xy = 1. Hence either A is cyclic or dihedral. Same conclusion holds for B too. Also similar conclusion 2 holds when s2 = y (since x = ab = 1).
−1 −1 −1 −1 3. When s2 = y and so t3 = a , then ac = xz = 1. In particular B is ±1 ±1 cyclic. If s 6= z, then t1 = b or t6 = b , and A is either cyclic or dihedral −1 (given also that the region containing t3(= c ) contains a corner labelled b).
So we suppose that s = z and so B = Z2. Hence no B-region is a 3-gon. So again A is cyclic or dihedral.
Proposition 4.4.15. Configuration (4, 2, 3, 1, 2) does not occur.
91 Chapter 4: One-relator product of two non-trivial groups with short relator
Figure 4.23: Configuration (4, 2, 3, 1, 2)
2 Proof. From the 4-zone we get the relations xy = b = 1. Also s1z = 1 and tiac = 1, where i ∈ {2, 3}. If the 3-zone joins vertices of opposite signs, then either x = y, b = c or x = z, b = a. In the first case A is cyclic and B is cyclic or a quotient of
D∞. In the second case both A and B are cyclic. Suppose the 3-zone joins vertices of same sign. If xz = 1 = bc, then both A and B 2 are cyclic. If x = 1 = ab, then A cyclic and B is either cyclic or a quotient of D∞. Finally if the 3-zone gives the relations xy = b2 = 1, then we can assume that the
B-region is not a 3-gon. This forces s1, s2 6= z, and hence both A and B are cyclic 2 −1 (since t2 = b or t3 = b). Otherwise z = 1 and a = c (since t2 = a and t3 = c). Then the result follows from Lemma 4.4.2.
Lemma 4.4.16. Configurations of Type 5 do not occur.
The proof of Lemma 4.4.16 is divided into four propositions, one for each configu- ration in Type 5. In each case we can assume by minimality and Lemma 4.4.2 that ±1 ±1 t1 ∈ {b , c }. In particular the following holds:
2 1. If t1 = b, then x = ab = 1, s2 = z and the region containing t1 is not a 3-gon. Also by Lemma 4.4.2 we assume yz 6= 1.
−1 −1 −1 2. If t1 = b , then xy = bc = 1 and s1 = z .
2 3. If t1 = c, then then xy = b = 1, s1 = x and the region containing s1 is not a 3-gon. Also by Lemma 4.4.2 we assume ac 6= 1.
−1 −1 −1 −1 4. If t1 = c , then xz = ab = 1 and s1 = x .
Proposition 4.4.17. Configuration (3, 3, 2, 2, 2) does not occur.
92 Chapter 4: One-relator product of two non-trivial groups with short relator
Figure 4.24: Configuration (3, 3, 2, 2, 2)
Proof. 1. Suppose that A is neither cyclic nor dihedral.
2 (a) If t1 = b, then we can assume that s2 = z, so that c = 1. Either A is
cyclic or s2 = z and so yz = 1. In either case we get a contradiction to assumption.
−1 2 −1 (b) If t1 = b , then we can assume that t2 = b and c = 1 or t2 = c . Since
at least one of t3t2a = 1 or t1t8a = 1 holds, then A is cyclic.
2 (c) If t1 = c, then it follows that A is cyclic or dihedral if c = 1 using ±1 t7t6b = 1. Otherwise if c = a , we use t5t4b = 1 to show A is cyclic.
−1 2 (d) If t1 = c , then we can assume that t2 = b and so c = 1. Since at least
one of t7t6c = 1 or t8at1 = 1 holds, we conclude that A is cyclic.
2. Suppose that B is neither cyclic nor dihedral.
2 (a) If t1 = b, then we can assume that s2 = z, so that c = 1. By Lemma ±1 −1 −1 4.4.2 and minimality we can assume t2 ∈ {c , b }. If t2 = b , then −1 −1 xz = 1 and so 1 = s1ys2 = zyz . This contradicts the assumption 2 −1 that y 6= 1. If t2 = c, then z = 1 and s2 = x. Similarly f t2 = c , then −1 −1 yz = 1 and so s2 = x . In both cases it follows that B is cyclic or
dihedral since zys2 = 1.
−1 2 (b) If t1 = b , then we can assume that t2 = c and so z = ac = 1, s2 = x. −1 Since t3t2a 6= 1, it follows that B is cyclic as 1 = s1ys2 = z yx.
2 (c) If t1 = c, we conclude from the other 3-zone that B is cyclic or ac = z = 1. In the latter, the result follows from Lemma 4.4.2.
−1 2 (d) If t1 = c , then we can assume that t6 = b and so y = 1. Also we can −1 −1 −1 −1 assume that t2 = b , so xz = ac = 1 and s2 = z . Hence B is
cyclic since s2ys2 = 1 (as t3t2a 6= 1).
93 Chapter 4: One-relator product of two non-trivial groups with short relator
Proposition 4.4.18. Configuration (3, 2, 3, 2, 2) does not occur.
Figure 4.25: Configuration (3, 2, 3, 2, 2)
Proof. 1. Suppose that A is neither cyclic nor dihedral.
2 (a) If t1 = b, we assume that s3 = y, so that c = 1. We then conclude from
t3t4c = 1 that A is cyclic. −1 2 (b) Suppose t1 = b . We can assume that t2 = c, so that a = xz = 1 and −1 s4 = x. It follows that s4s3z 6= 1. Hence 1 = t1t6a = b t6a, and so A is cyclic.
(c) If t1 = c, we conclude from the 2-zone with corner labelled c that either 2 ±1 c = 1 or c = a . It follows from 1 = t1t6a = ct6a that A is cyclic or dihedral.
−1 2 (d) If t1 = c , then we can assume that s4 = x so that t2 = c, xz = a = 1 −1 and s4s3z 6= 1. It follows from 1 = t1t6a = c t6a that A is cyclic.
2. Suppose that B is neither cyclic nor dihedral.
±1 ±1 (a) If t1 = b, then we can assume that x = y (alternatively x = z ) since
t2 6= a. It follows that B is cyclic or dihedral (using the 2-zone containing y (alternatively z).
−1 2 (b) If t1 = b , then we can assume that t6 = c and so z = 1. Since at least
one of s1ys2 = 1 or s4s3z = 1 holds, B is cyclic. 2 (c) If t1 = c, then we can assume that t6 = c and so z = 1. It follows that
s4s3z = 1 (since s1ys2 6= 1). Hence B is cyclic. −1 2 (d) If t1 = c , we can assume that y = 1 (using the 2-zone with corner −1 −1 labelled y). Also we can assume that t5 = a , t6 = b and so t6at1 6= 1. −1 It follows from 1 = s1ys2 = x ys2 that B is cyclic.
94 Chapter 4: One-relator product of two non-trivial groups with short relator
Proposition 4.4.19. Configuration (3, 3, 3, 2, 1) does not occur.
±1 Proof. By Lemma 4.4.2 and minimality we can assume t2 6= a .
Figure 4.26: Configuration (3, 3, 3, 2, 1)
1. Suppose that A is neither cyclic nor dihedral.
2 (a) If t1 = b, we can assume that t2 = b, so that yz = c = 1. The rest of the result follows from Lemma 4.4.2.
−1 (b) Suppose t1 = b , then A is cyclic since at least one of t1t4a = 1 or
t2at3 = 1 holds.
(c) If t1 = c, then using the 2-zone with corner labelled c we conclude that 2 ±1 ±1 c = 1 or c = a or c = b . Since t4at1 = t3t2a = 1, we have that A is cyclic or dihedral.
−1 2 (d) If t1 = c , we assume that t2 = b, so that c = zy = 1. Hence s1ys2 6= 1. −1 So t4ac = 1 and hence A is cyclic.
2. Suppose that B is neither cyclic nor dihedral.
±1 ±1 (a) If t1 = b, then by minimality and Lemma 4.4.2 s2 ∈ {y , z }. Hence −1 2 ±1 z = y or z = 1 and 1 = s1ys2 = zyx . So B is cyclic or dihedral.
−1 2 (b) If t1 = b , we can assume that t2 = c and so z = 1, t2at3 6= 1. It follows
that s1ys2 = 1 and so B is cyclic.
(c) If t1 = c, we can assume that t2 = c. Hence the result follows from Lemma 4.4.2 since ac = 1.
−1 (d) Suppose that t1 = c . Since at least one of s1ys2 = 1 or s4s3y = 1 holds,
we conclude that B is cyclic unless s1ys2 6= 1 and s4 = s3 = y. But in this case bc = xz = 1 and we can apply Lemma 4.4.2.
95 Chapter 4: One-relator product of two non-trivial groups with short relator
Proposition 4.4.20. Configuration (3, 3, 1, 3, 2) does not occur.
Figure 4.27: Configuration (3, 3, 1, 3, 2)
Proof. 1. Suppose that A is neither cyclic nor dihedral.
(a) If t1 = b, then using ct = 1 (from the other 3-zone) and t5t6c = 1 we conclude that A is cyclic.
−1 2 2 (b) Suppose t1 = b . If c = 1 (alternatively b = 1), we conclude that A is
cyclic since at least one of t1t6a = 1 or t4ct5 = 1 holds. Hence we assume ±1 −1 that s2 = x , and so t2 = c . But also at least one of t1t6a = 1 or
t2at3 = 1 holds. Hence A is again cyclic.
(c) If t1 = c, then we can assume that t2 = c and so the result follows from Lemma 4.4.2.
−1 −1 (d) If t1 = c , then we can assume that s4 = c. Since x yx = s1ys2 6= 1,
we have 1 = at3t2 = at3c, and so A is cyclic.
2. Suppose that B is neither cyclic nor dihedral.
2 (a) If t1 = b, we conclude from one of the other 3-zones that z = 1 or ±1 z = y . Either way it follows from 1 = s1ys2 = s4s3x that B is cyclic or dihedral.
−1 2 (b) If t1 = b , then we can assume that t2 = c, so z = 1 and s2 = x. Since
t3t2a 6= 1, we have that s1ys2 = 1 and so B is cyclic.
(c) If t1 = c, we can assume that t2 = c and so the result follows from Lemma 4.4.2.
96 Chapter 4: One-relator product of two non-trivial groups with short relator
−1 2 2 (d) If t1 = c , we can assume that t4 = b and so y = 1. Since z 6= 1 by −1 −1 assumption, s2 = z and so ac = 1. It follows that s1ys2 = 1 (since
t3t2a 6= 1). Hence B is cyclic.
Part 2
Here we consider the bad cases and show that part 2 of Theorem 4.4.1 holds. We do this in number of Lemmas, one for each of the three configurations.
Lemma 4.4.21. Configuration (4, 2, 2, 2, 2) does not occur.
Proof. Configuration (4, 2, 2, 2, 2) is depicted in Figure 4.28. From the 4-zone we 2 get the relations xy = b = 1. Clearly, this implies B is cyclic whenever s1, s2 6= z. We can assume without loss of generality that the two regions with corners labelled t1 and t2 are 3-gons.
Figure 4.28: Configuration (4, 2, 2, 2, 2)
In other words t1ac = s1z = t2t3c = 1. There are four cases to consider depending −1 −1 on the value of s1. Note that s1 6= y (as that will imply that t1 = c and so a = 1).
−2 ±1 1. Suppose s1 = x and hence t1 = a and t2 = b. Then c = a and t3 ∈ {a , b, c}. In each case A is cyclic.
−1 −1 −1 2. Suppose s1 = x , and hence t1 = b and t2 = a . Then b = ac and a = t3c. −1 If t3 6= b, then A is cyclic. Otherwise if t3 = b, then b = c a. It follows that 2 −1 2 a = (ac)(c a) = b = 1. So A is a quotient of D∞.
2 3. Suppose s1 = y and hence t1 = b and t2 = c. Then bac = 1 = t3c . If t3 6= c, 3 then A is cyclic. So we suppose that t3 = c, hence c = 1 and t4 = a. At least
97 Chapter 4: One-relator product of two non-trivial groups with short relator
one of t5ba = 1 or act8 = 1 holds. In the former, we can assume that t8 = b (as −1 2 otherwise A is cyclic). It follows that s2 ∈ {x, y }, and so z = 1. Similarly
in the latter, we can assume that t5 = c, and so from the 2-zone we have that xz = 1. Again z2 = 1. This completes the proof.
Lemma 4.4.22. Configuration (4, 2, 4, 1, 1) does not occur.
Proof. Without loss of generality, Figure 4.29 shows ϑ with configuration (4, 2, 4, 1, 1). The 4-zone gives the relations xy = b2 = 1.
Figure 4.29: Configuration (4, 2, 4, 1, 1)
Suppose s1 = x, and so t1 = a and t2 = b. If the region containing s2 is not a 3-gon, then all other regions which are not 2-gonal are 3-gons. In particular 2 t2ac = bac = 1 = t1ac = a c. So A is cyclic. Hence we suppose the region −1 containing s2 is a 3-gon, so s2s3z = 1. Note that t4 6= b (as otherwise s3 ∈ {x, y } and so s2 = 1). So act4 = 1 = bac, where t4 ∈ {a, c}. Again A is cyclic.
−1 −1 −1 Suppose s1 = x , and so t1 = b and t2 = a . In particular the region containing t2 is not a 3-gon. Hence all other regions which are not 2-gonal are 3-gons. So −1 t1ac = bac = 1 = s2s3z. Note that t3, t4 6= b (since z = x = y ). It follows that t3ac = 1 with t3 ∈ {a, c}. Hence A is cyclic.
±1 By symmetry, similar argument takes care of the cases where s1 = y . Hence we must have that s1 = z. This completes the proof.
Lemma 4.4.23. Configuration (4, 2, 4, 2) does not occur.
Proof. Without loss of generality, Figure 4.30 shows ϑ with configuration (4, 2, 4, 2). If any of the two 4-zones joins vertices of opposite signs, then in particular B is cyclic. Also either b = a (alternatively b = c). We can use the four A-regions to show that A is cyclic or dihedral, hence the result.
98 Chapter 4: One-relator product of two non-trivial groups with short relator
Figure 4.30: Configuration (4, 2, 4, 2)
2 So we assume that t2 = t6 = a, t1 = t5 = c. Hence xy = b = 1. If s1 = z or s2 = z or more generally B is Z2, then we are done. Suppose not, then {s1, s2} ∈ {{x, y−1}, {x−1, y}, {y}, {x}}. At this point we amend the picture M to get Mˆ in which the number of 2-gonal A-regions is maximal. We can assume that the only positively-curved interior vertex in Mˆ has configuration (4, 2, 4, 2); and proceed as before. Since there are two A-regions with consecutive corners labelled b±1, a, c, we have that either bac = 1, or tbac = 1 and there is a relation of length at most 3 involving only a and c. In the latter A is cyclic. In the former r has order 2 for otherwise its image under the projection map onto A is abc. Hence abc = acb = 1, hence A is abelian. In both case G satisfies the Freiheitssatz. This completes the proof.
This shows that our assumption that Theorem 4.4.1 fails was false, hence the proof of the Theorem 4.3.1 is complete.
4.5 Relator of length four with power n ≥ 2
In the section we consider the one-relator group
(A ∗ B) G = , (4.14) N(rn) where n ≥ 2 and r = abcd with a, c ∈ A and b, d ∈ B.
Theorem 4.5.1. Let G be as in 4.13. Then the following holds for G.
1. Freiheitssatz holds for G.
2. The word r has order n in G.
Lemma 4.5.2. If b = d−1 or a = c−1, then Theorem 4.14 holds.
99 Chapter 4: One-relator product of two non-trivial groups with short relator
Proof. Without loss of generality, we assume that b = d−1. If n ≥ 3, then the result holds by [62]. Hence we assume that n = 2. By assumption, A is not cyclic. Suppose Freiheitssatz fails. We construct a W-minimal spherical picture M over G on D2. The label of each vertex contains two of letter a and two of letter b. Hence by Lemma 4.2.5, at least one of the interior A-regions must have degree 2 or 3. Since A is not cyclic, any relation of length 2 or 3 between a and c must be a power of a or a power of c. Suppose that a has order p ∈ {2, 3} and c has order q ≥ 2. Then in particular hai ∩ hci = {1} in A.
i −i The normal closure N of A in G is the quotient of m copies Ai := b Ab of A n (where m is the order of b in B) by the relators (aici+1) (subscripts mod m). n Let Hi := hAi, ai−1 | (ai−1ci) = 1i, the free product of Ai with a triangle group p q n Ti := hai−1, ci | ai−1 = ci = (ai−1ci) = 1i, amalgamated over the cyclic subgroup hcii.
If m = ∞ then G is the HNN extension of H0 with associated subgroups ha0i and ha−1i and stable letter b, and the result clearly holds in this case. Suppose then that m < ∞. By [[88], Theorem 7], Hi acts on a tree X with fundamental domain a segment T (see Figure 4.31) with Ai, Ti and hcii the stabilizers of the vertices u, v
Figure 4.31: A segment T of X and the edge e respectively. The edge e partitions X into two disjoint components:
Xu (the component containing the vertex u) and Xv (the component containing the j vertex v). Since hai−1i∩hcii = {1} in Ti and haii∩hcii = {1} in Ai , ai Xu ⊆ Xv and j ai−1Xv ⊆ Xu for j 0 mod p (i.e no non-trivial element of hai−1i or haii fixes the e).
It follows from the Ping-Pong Lemma that the subgroup Fi < Hi generated by ai−1 0 and ai is the free product hai−1i ∗ haii. Let H := b ∗ha1i · · · ∗ham−2i Hm−1. Then the 0 0 subgroup of H generated by am−1 and a0 is isomorphic to F0, and N = H0 ∗F0 H .
Hence A = A0 < H0 < N < G. Trivially B < G. Suppose that r does not have order n in G. The proof is similar to that in Lemma 4.4.2. We construct a W-minimal spherical clique-picture M over G. A typical clique-label will have the form
aα1 bcβ1 b−1 . . . aαk byβk b−1.
100 Chapter 4: One-relator product of two non-trivial groups with short relator
G is induced by the generalised triangle group
T (p, q, n) = ha, c | ap, cq, (ac)2i. (4.15)
Hence by the Spelling theorem for generalised triangle groups, it follows that for an interior clique-label k ≥ 2. Also we assume that A is not cyclic. In any region of M, we may assume that no two consecutive corner labels are of the form aα or cβ (for otherwise we could amalgamate the two cliques into one, forming a smaller clique-picture). In particular, if a 2-zone is labelled by aα at one corner, the opposite corner has to be a c±1. But also this can not happen since from the proof of Freiheitssatz, p = 2 or 3. Hence any A-region has degree at least four. It follows that M has at most one clique of positive curvature, namely the clique corresponding to the exceptional vertex. Hence the total curvature of M is less than 2π. This contradiction implies that r has order n. This completes the proof.
Corollary 4.5.3. M has a positively-curved interior vertex.
Lemma 4.5.4. If a 3-zone is incident at ϑ, then Theorem 4.14 holds.
Figure 4.32
Proof. Suppose that two vertices u and v of M are joined by the 3-zone. Without loss of generality we can assume that the corner labels at u are a and b. Sup- pose x and y are the corresponding labels at v i.e ax = 1 = by. Then (x, y) ∈ {(a, d), (c, b), (c−1, d−1)}. In each case, the result holds by assumption and Lemma 4.5.2.
Proof of Theorem 4.5.1. By Lemma 4.5.2 we can assume that neither b = d−1 nor a = c−1 holds. Suppose that Theorem 4.5.1 fails. Then we obtain a obtain a spherical diagram M over G with boundary label w ∈ A ∪ B. Each vertex has 4n (n ≥ 2) edges incident on it and with label r±n. By Lemma 4.2.5, M has a positively curved interior vertex
101 Chapter 4: One-relator product of two non-trivial groups with short relator
4n ϑ. It follows from Lemma 4.5.4 that 4n − 5 ≤ i.e n = 2. This agrees with the 2 already known result that Freiheitssatz holds for n ≥ 4 and proves it also for n = 3. When n = 2, it follows from Lemma 4.5.4 that it is enough to consider the cases where the number of regions of v of degree 2 is three or four. If there are four regions of degree 2, then without loss of generality there are two A-regions –say ∆1, ∆2 each of degree 2 separated by a B-region of degree 3.
Figure 4.33: A positively-curved interior vertex
Without loss of generality, the possible cases for the labels of the two A-vertices are:
2 2 1.∆ 1 has label a and ∆2 has label c ;
2 −1 2.∆ 1 has label a and ∆2 has label ca ;
−1 −1 3.∆ 1 has label ac and ∆2 has label ca .
In all cases, A has faithful representation in PSL2(C). Also B is cyclic as the label of the B-region contains both the elements b and d. This contradicts the starting hypothesis and hence the proof. If on the other hand there are three 2-zones, then v has degree 5. Hence v has at least four regions of degree 3. If there is a region of degree 3 separating two 2-zones, the the result follows as before. Without loss of generality, we can assume that the three 2-zones have labels ac, ca and b2 respectively as shown in Figure 4.33. Hence a = c = 1, contradicting hypothesis.
102 Chapter 4: One-relator product of two non-trivial groups with short relator
4.6 Special case of n = 1 and relator has length four.
In the section we consider the one-relator group
(A ∗ B) G = , (4.16) N(r) where r = abcd with ha, ci = A0 ≤ A and hb, di = B0 ≤ B. By putting restrictions on the relator and on the order of the factors, Shwartz [90] has some classification on when Freiheitsatz holds. For us the only requirement is that each of A0 and B0 is a two-generator groups with faithful representation in PSL2(C).
Theorem 4.6.1. Let G be as in 4.16. Under above assumption, Freiheitssatz holds for G.
We make the following observation about representation of groups in SL2(C). For more details see ([16],[33],[3]). Let F2 = hx, yi be a free group of rank two. Then the irreducible representations ρ : F2 −→ SL2(C) are parametrised (up to conjugacy) by the three independent parameters namely Trρ(x), Trρ(y) and Trρ(xy). In other words, the character variety of conjugacy classes of irreducible representations ρ is an affine algebraic set contained in C3. Recall that ρ is reducible if and only if Tr[ρ(x), ρ(y)] = 2. This criterion translates to
Tr(ρ(x))2 + Tr(ρ(y))2 + Tr(ρ(xy) − Tr(ρ(x)Tr(ρ(y)Tr(ρ(xy) = 4. (4.17)
So the irreducible character variety is the complement of a cubic surface in C3.
If we rather consider the free group in three variables F3 = hx, y, zi, then by [33] a conjugacy class of irreducible representations ρ : F3 −→ SL2(C) is determined by seven parameters: Trρ(x), Trρ(y), Trρ(z), Trρ(xy), Trρ(xz), Trρ(yz) and Trρ(xyz) which are no longer independent but satisfy the following polynomial equation.
Trρ(x)2 + Trρ(y)2 + Trρ(z)2 + Trρ(xy)2 + Trρ(xz)2 + Trρ(yz)2 + Trρ(xyz)2
+Trρ(xy)Trρ(xz)Trρ(yz) − Trρ(x)Trρ(y)Trρ(xy) − Trρ(x)Trρ(z)Trρ(xz)
−Trρ(y)Trρ(z)Trρ(yz) + Trρ(xyz)Trρ(x)Trρ(y)Trρ(z) − Trρ(xyz)Trρ(x)Trρ(yz)
−Trρ(xyz)Trρ(y)Trρ(xz) − Trρ(xyz)Trρ(z)Trρ(xy) = 4 (4.18)
The irreducible character variety is 6-dimensional, and therefore comprised of a
103 Chapter 4: One-relator product of two non-trivial groups with short relator hyper-surface in C7 defined by Equation 4.18, minus the subset corresponding to reducible representations, which is the intersection the three cubic curves arising from the commutators Tr[ρ(x), ρ(y)], Tr[ρ(x), ρ(z)] and Tr[ρ(y), ρ(z)], where [x, y] = xyx−1y−1. Using above comments we give a proof of Theorem 4.6.1.
Proof of Theorem 4.6.1. Let X,Y and Z be variable matrices in SL2(C). The aim of the proof is to show that one can choose X, Y and Z such that a 7→ X, c 7→ Z −1 gives a faithful representation A 7→ PSL2(C) and b 7→ Y , d 7→ (XYZ) gives a faithful representation B 7→ PSL2(C).
Such a triple of matrices is a representation of the free group F3 of rank 3. Recall that the character variety of representations F3 → PSL(2, C) is given by the seven parameters Tr(X), Tr(Z), Tr(Y ), Tr(XY ), Tr(XZ), Tr(YZ) and Tr(XYZ) subject to a single polynomial equation
Tr(X)2 + Tr(Y )2 + Tr(Z)2 + Tr(XY )2 + Tr(XZ)2 + Tr(YZ)2
+ Tr(XYZ)2 + Tr(XY )Tr(XZ)Tr(YZ)
− Tr(X)Tr(Y )Tr(XY ) − Tr(X)Tr(Z)Tr(XZ) − Tr(Y )Tr(Z)Tr(YZ) (4.19)
+ Tr(X)Tr(Y )Tr(Z)Tr(XYZ) − Tr(X)Tr(YZ)Tr(XYZ)
− Tr(Y )Tr(XZ)Tr(XYZ) − Tr(Z)Tr(XY )Tr(XYZ) = 4
By hypothesis, faithful representations of A and B in PSL2(C) exist. Moreover they are parametrised by fixing suitable values for Tr(X), Tr(Z), Tr(XZ), Tr(Y ), Tr(XYZ) and Tr(XYZY −1). By the repeated application of trace relation
Tr(MN) = Tr(M)Tr(N) − Tr(MN −1) (4.20) for arbitrary matrices M and N can write:
Tr(XYZY −1)=Tr(Y )Tr(XYZ)−Tr(XY )Tr(YZ)+Tr(X)Tr(Z)−Tr(XZ). (4.21)
Hence if we fix suitable values for Tr(X), Tr(Y ), Tr(Z), Tr(XZ) and Tr(XYZ), we have two free variables α := Tr(XY ) and β := Tr(YZ) which are required to satisfy the quadratic equation which fixes the value of
Tr(XYZY −1) = Tr(Y )Tr(XYZ) − αβ + Tr(X)Tr(Z) − Tr(XZ). (4.22)
104 Chapter 4: One-relator product of two non-trivial groups with short relator
Combining this with Equation (4.19), and fixing Tr(X), Tr(Z), Tr(XZ), Tr(Y ), Tr(XYZ), we have a pair of quadratic equations in α, β of the form
αβ = c1 (4.23) and 2 2 α + β + c2αβ + c3α + c4β = c5 (4.24) for suitable constants c1, . . . , c5. It is routine to check that any such pair of equations can be solved in C. Any solution gives a representation ha, b, c, di −→ SL2(C) that induces the given faithful representations of A and B in PSL2(C) up to conjugacy, mapping the word abcd to the identity element. This completes the proof.
105 Chapter 5
One-relator product of three non-trivial finite cyclic groups
5.1 Preamble
We prove a very important result about one relator free product of cyclic groups. Let (G ∗ G ∗ G ) G = a b c N(w) p q r where Ga = ha | a i, Gb = hb | b i, and Gc = hc | c i. We show that each of Ga,
Gb and Gc embeds in G in the case where w has non-zero exponent sum in each of the generators (modulo their respective orders). In [58], it was shown that such a group in non-trivial. Indeed the proof there shows that each of Ga, Ga and Ga embeds in G in the case where p, q and r are distinct primes. We will adapt the methods in [58] to show that a stronger conclusion holds for arbitrary non-zero triple p, q, r. More precise we show that in a one-relator product of three non-trivial cyclic groups, at least one of the factors is a Freiheitssatz factor. Before that we present some preliminary results most of which are based from [58].
5.2 S1-equivariant homotopy and degree of maps
Definition 5.2.1. A homotopy between two continuous maps f1, f2 : X −→ Y is a continuous map H : X × [0, 1] −→ Y such that, if x ∈ X then H(x, 0) = f1(x) and
H(x, 1) = f2(x). If such H exist, we say that f1 and f2 are homotopic.
Definition 5.2.2. Let X and Y be G-spaces for some group G. Then a continuous map f : X −→ Y is said to be G-equivariant if f(gx) = gf(x) for each x ∈ X and g ∈ G.
Definition 5.2.3. A G-equivariant homotopy between two G-equivariant maps
106 Chapter 5: One-relator product of three non-trivial finite cyclic groups
f1, f2 : X −→ Y is a G-equivariant map H : X × [0, 1] −→ Y , where G acts on X × [0, 1] by g(x, t) = (gx, t), so that H(x, 0) = f1(x) and H(x, 1) = f2(x). If such H exist, we say that f1 and f2 are G-equivariantly homotopic.
Definition 5.2.4. A G-equivariant map f1 : X −→ Y is a G-equivariant homotopy equivalence if there is a G-equivariant map f2 : Y −→ X so that f1f2 and f2f1 are G-equivariant homotopic to the respective identity maps. In which case we say that X and Y are G-equivariant homotopy equivalent.
In what follows, G will be S1 and the G-spaces will be S2 and S3. Later on we shall make precise what the S1-action is. Recall that for some real number r and integer n, an n-sphere of radius r is the set
n n+1 2 2 2 2 S = {(x1, x2, . . . , xn+1) ∈ R | x1 + x2 + ... + xn+1 = r }. (5.1)
We shall be mostly interested in the unit n-sphere i.e r = 1. Identify S3 with the group of unit quaternions
3 2 2 2 2 S = {x1 + x2i + x3j + x4k | x1 + x2 + x3 + x4 = 1}, (5.2) S2 with the subset
2 2 2 2 S = {x2i + x3j + x4k | x2 + x3 + x4 = 1}, (5.3) and S1 with the subgroup
1 2 S = {x1 + x2i | x1 + x2 = 1}. (5.4) We shall use Sn and SU(n) interchangeably. For example, by the trace of
3 L = x1 + x2i + x3j + x4k ∈ S , (5.5) we mean the trace of the corresponding matrix in SU(2). That is the trace of
! x + x i x + x i L = 1 2 3 4 . (5.6) −x3 + x4i x1 − x2i
Definition 5.2.5. The degree of a continuous map f : Sn −→ Sn is α ∈ Z if the n n induced map f∗ : Hn(S ) −→ Hn(S ) is a multiplication by α, where Hn(.) stands for the n-th homology group of Sn.
Lemma 5.2.6 ([58] Lemma 2.1). Let X be a simply-connected space equipped with an S1-action, and let f, g : S2 −→ X be S1-equivariant maps. Then f is equivariantly
107 Chapter 5: One-relator product of three non-trivial finite cyclic groups homotopic to g if and only if there are paths in the fixed subspace joining f(+i) to g(+i), and f(−i) to g(−i).
Corollary 5.2.7 below gives us a way of computing the degree of S1-equivariant maps from S2 to itself.
Corollary 5.2.7 ([58] Corollary 2.2). The degree of any S1-equivariant map f : S2 −→ S2 is one of the following:
1. 0 if f(i) = f(−i) ∈ {±i};
2. +1 if f(i) = i and f(−i) = −i; or
3. −1 if f(i) = −i and f(−i) = i.
Lemma 5.2.8. Two elements in S3 are conjugates if and only if they have same eigenvalues.
Proof. The forward direction is clear. For the reverse, let T ∈ S3. It follows from the ! λ 0 spectral theorem that STS−1 = for some S ∈ U(2) and λ an eigenvalue 0 λ¯ of T . The final step is obtained by replacing S withsS ¯ where s = pdet(S). SincesS ¯ ∈ S3 and eigenvalues are completely determined by the trace, the result follows.
Lemma 5.2.9. The conjugacy classes of S3 are the sets C(θ) = {cos θ +sin θvv | v ∈ S2} where 0 ≤ θ ≤ π.
3 Proof. We begin by showing that any element L = x1 + x2i + x3j + x4k in S can be expressed in the form cos θ + sin θvv for some 0 ≤ θ ≤ π and v ∈ S2. Since −1 −1 ≤ x1 ≤ 1, take θ = cos (x1) and v = (x2i + x3j + x4k)/ sin θ. Note that L has trace 2x1.
p 2 The eigenvalues of L are x1 ± x1 − 1, so depends just on x1 as does θ. The result follows from Lemma 5.2.8.
Now we are in a position to describe S1 action on S3 and S2. S3 is a group and so acts on itself by conjugation. The fixed set of the S3-action consists of precisely C(θ) for 0 ≤ θ ≤ π, each of which is a topological 2-sphere with the exception of C(0) = {1} and C(π) = {−1}. By restricting to the subgroup S1, we get an action of S1 on S3 and on S2. The fixed set of S1-action on S3 is itself, the fixed set of S1 action on S2 is S1 ∩ S2 = {±i}.
Lemma 5.2.10. The spaces S2 and S3 − {±1} are S3-equivariant homotopy equiv- alent.
108 Chapter 5: One-relator product of three non-trivial finite cyclic groups
Proof. The inclusion map ι : S2 → S3 − {±1} is an S3-equivariant map. It has an inverse ψ : S3 − {±1} → S2, sending cos θ + sin θvv to v, which is also S3-equivariant map.
What we really need are the maps ι and ψ.
5.3 Main result
In this section, we prove the main result as a corollary to the following Lemma.
p q r Lemma 5.3.1. Let Ga = ha | a i, Gb = hb | b i and Gc = hc | c i be cyclic groups, with p, q, r are prime powers.. Let w ∈ Ga ∗ Gb ∗ Gc be an element such that the exponent sums of a, b, c in w are not divisible by p, q, r respectively (equivalently, w is not contained in the normal closure of any two of Ga,Gb,Gc). Then each of
Ga,Gb,Gc embeds via the natural map into
(G ∗ G ∗ G ) G = a b c . N(w)
Proof. We know the result holds when p, q, r are primes by [[58], Theorem 4.1]. We suppose that atleast on of p, q, r is a prime power but not prime. Let F (a, b, c) denote the free group of rank 3 with generating set {a, b, c}. The aim is to produce a representation of F (a, b, c) in H (where H denotes the quaternions) such that a, b, c have orders p, q, r respectively and w ∈ F (a, b, c) is sent to the identity. Suppose that n is the exponent sum of a in w. The assumption that n is not divisible by p implies that n = tp + s with 0 < s < p. By replacing w with wa−tp ∈ F (a, b, c) which leaves G unchanged, we can always assume that n < p. If m is co-prime to p m m then a 7→ a induces an automorphism of Ga. Thus, replacing a by a in w gives 0 a new word w ∈ Ga ∗ Gb ∗ Gc such that the resulting group
(G ∗ G ∗ G ) G0 = a b c N(w0)
0 is isomorphic to G (and such that Ga embeds in G if and only if it embeds in G). Moreover, the exponent sum of a in w0 is mn. By Bezout’s Lemma we may choose m such that mn ≡ gcd(n, p) mod p. Thus without loss of generality we may assume that n divides p (and similarly the exponent sums of b, c in w divide q, r respectively). Now suppose that p and n (the exponent sum of a in w) are powers of a prime τ – say p = τ t and n = τ s where 0 ≤ s < t. If τ is an odd prime, define
(τ t−s − 1)π θ = . (5.7) p 2τ t 109 Chapter 5: One-relator product of three non-trivial finite cyclic groups
If τ = 2, define
(2t−s−1 − 1)π θ = (5.8) p 2t π unless s = t − 1, in which case define θp = 2 . Recall that an element cos(θp) + t−1 v sin(θp) ∈ SO(3) has order p if and only if θp is a multiple of π/p but not π/τ , for any vector v ∈ S2. Hence for any vector v ∈ S2, the map
αv : Ga → H, a 7→ cos(θp) + v sin(θp),
3 ∼ induces a faithful representation Ga → S /{±1} = SO(3). Moreover, the real part n π π of αv(a ) is cos(ψp) where ψp := nθp and 4 ≤ ψp ≤ 2 .
Similarly, we can define maps βv : Gb → H and γv : Gc → H that induce faithful representations Gb,Gc → SO(3), and such that, if e, f denote the exponent-sums of e f b, c in w, then the real part of βv (b ) is cos(ψq) and the real part of γv (c ) is cos(ψr) π π where ψq, ψr ∈ [ 4 , 2 ].
The numbers ψp, ψq, ψr satisfy a triangle inequality: none is greater than the sum of the other two. Hence, for example, the triple (αi, βi, γ−i) induces a homomorphism
3 δ : Ga ∗ Gb ∗ Gc → S
3π that sends w to cos(θ) + i sin(θ) with 0 ≤ θ ≤ 4 . If θ = 0 then δ induces a representation G → SO(3) that is faithful on each of Ga,Gb,Gc and we are done. So assume that θ > 0. In other words the imaginary part of δ(w) is greater than
0. Similar remarks apply to the triples (αi, β−i, γi) and (α−i, βi, γi). Hence also the triple (αi, β−i, γ−i) induces a representation δ with the imaginary part of δ(w) less 2 3 1 than 0. The map S → S , v 7→ (αi, β−i, γv )(w) is an S -equivariant map under the conjugation action. It follows that it either sends some v to ±1 ∈ S3 (in which case
(αi, β−i, γv ) gives a representation G → SO(3) that is faithful on each of Ga,Gb,Gc), 3 or represents +1 in H2(S \ {±1}) (isomorphic to Z), the second homology group of S3 \ {±1} by Lemma 5.2.6.
2 3 Similarly, the map v 7→ (αi, βi, γv )(w) either maps some v ∈ S to ±1 ∈ S and so gives a representation G → SO(3) that is faithful on each of Ga,Gb,Gc, or represents 3 ∼ 2 one of 0, −1 ∈ H2(S \ {±1}) = Z. Now any path P : [0, 1] → S from −i to i gives rise to a homotopy
t 7→ v 7→ (αi, βP (t), γv)(w) between the above two maps. If (αi, βP (t), γv )(w) 6= ±1 for all t and for all v, then we can regard this as a homotopy of maps S2 → S3 \{±1}. This is a contradiction since
110 Chapter 5: One-relator product of three non-trivial finite cyclic groups
3 the two maps belong to different homology classes in H2(S \{±1}). Hence for some t and some v, the map (αi, βP (t), γv ) sends w to ±1 and so induces a representation
G → SO(3) that is faithful on each of Ga,Gb and Gc.
It follows that each of the natural maps from Ga,Gb and Gc to G is injective, as required.
The general case of the Freiheitssatz for G follows from the special case of prime powers together with the Chinese Remainder Theorem by an easy induction.
p q r Theorem 5.3.2. Let Ga = ha | a i, Gb = hb | b i and Gc = hc | c i be cyclic groups.
Let w ∈ Ga ∗Gb ∗Gc be an element such that the exponent sums of a, b, c in w are not divisible by p, q, r respectively. Then each of Ga,Gb and Gc embeds via the natural map into (G ∗ G ∗ G ) G = a b c . N(w) Proof. For the inductive step, suppose that p = mn with gcd(m, n) = 1. Since the exponent sum of the generator a in w is non-zero modulo p, we can assume that it is non-zero modulo m. Now factor out am and apply the inductive hypothesis. This shows that the maps Gb → H and Gc → H are injective. It also shows that the kernel K of Ga → H is contained in the subgroup hmi. If the exponent sum of a in w is also non-zero modulo n, then by interchanging the roles of m and n in the above we see that K is contained in hni. However, if the exponent-sum of a in w is divisible by n, then the same is automatically true: K is contained in hni. Finally, we know that K is contained in the intersection of hmi and hni. But this intersection is trivial by the Chinese Remainder Theorem, so we deduce that Ga → H is injective. Remark 5.3.3. It is clears that if a occurs in w with zero exponent sum modulo p, then A is a Freiheitssatz factor of G. Combining this Remark 5.3.3 with Theorem 5.3.2 we obtain: Corollary 5.3.4. In a one-relator product of three non-trivial cyclic groups, at least one of the factors is a Freiheitssatz factor. Definition 5.3.5. A group H is said to be finitely annihiliated if for every non- trivial element h ∈ H, there exist a finite index normal subgroup N of H such that h is trivial in H/N. In [13], Chiodo used the result of Howie [58] to show that the free product G of three cyclic groups of distinct prime orders is finitely annihiliated. The proof uses nothing more than the fact that finitely generated subgroups of SO(3) are residually finite. Hence our result extends this to the case where the cyclic groups are arbitrary. Corollary 5.3.6. Free product of three cyclic groups is finitely annihiliated.
111 Chapter 6
One-relator product of three non-trivial groups with short relator
6.1 Preamble
In Chapter 5 we considered a one-relator product of three finite cyclic groups. Here we do the same only that this time we allow the factor groups to be arbitrary. It is conjectured in [58] that such a group is non-trivial. We prove this conjecture in a special case. In particular we prove the following theorem.
Theorem 6.1.1. Suppose A, B and C are any three non-trivial groups. Then
(A ∗ B ∗ C) G = N(w) is non-trivial, for any word w ∈ A ∗ B ∗ C with `(w) < 9.
Theorem 6.1.1 holds trivially if w is in the normal closure of any of the factors. Hence we can assume by results in Chapter 4 that w contains at least two letters in each of the three factors. To see why this is true, suppose by way of contradiction that w contains one letter from A, say α. Then a cyclic conjugate of w has the form w = αW , where W ∈ B ∗ C and `(W ) ≤ 7 ( and so some cyclic conjugate of W has length at most 6). If α has order n in A, then we define the group
(B ∗ C) H = . N(W n) From Corollary 4.1.2 we know that W has order n in H. Hence
G = Ahαi ∗hW i H.
112 Chapter 6: One-relator product of three non-trivial groups with short relator
So G is non-trivial. In particular it is enough to consider the cases where the length of the relator is at least six and at most eight, as a cyclically reduced word in the free product A ∗ B ∗ C. We now prove some technical results from which Theorem 6.1.1 will follow as a corollary.
6.2 Technical Lemmas
We can assume that up to cyclic permutation w has the form w = c1Uc2V where c1, c2 ∈ C and U, V ∈ A ∗ B with `(U) + `(V ) ≤ 6. The group G is non-trivial if c1c2 = 1 or UV ∈ N(A)∪N(B). Hence we assume that neither of the two conditions holds. If without loss of generality we assume `(U) ≤ `(V ), then the possibilities for U and V as words in the free product A ∗ B are as follows:
1. U or V is in the normal closure of A or B;
±1 2. U = (αβ) and V ∈ {α1β1, α1β1α2, α1β1α2β2};
3. U = αβα1 and V ∈ {β1α2β2, α2β1α3} (by symmetry), where α (with or without subscript) is an element of A and β (with or without subscript) is an element of B. We show that in each of the possibilities listed above G is non-trivial.
Lemma 6.2.1. If U or V is in the normal closure of A or B, then G is non-trivial.
Proof. Without loss of generality, we can assume that U is in the normal closure of A. In other words, there exist a γ ∈ A ∗ B such that U = γ−1αγ. Hence
−1 w =c1γ αγc2V.
˜ −1 ˜ −1 −1 −1 Let C = γCγ and W =c ˜1αc˜2V = γc1γ αγc2γ γV γ . So G is trivial if and only if (A ∗ B ∗ C˜) G0 = (6.1) N(W ) is trivial. Hence it is enough to consider the case where U = α (i.e γ = 1).
−1 By assumption c1 6= c2 . Hence c1αc2 has infinite order in A ∗ C. It follows that the subgroup of A ∗ C generated by A and c1αc2 is isomorphic to the free product A ∗ Z. If A and V generates a subgroup of A ∗ B which is also isomorphic to A ∗ Z, then
G = (A ∗ B)hA,V i ∗hA,c1αc2i (A ∗ C). (6.2)
113 Chapter 6: One-relator product of three non-trivial groups with short relator
So in this case G is non-trivial.
Suppose then that hA, V i is not isomorphic to A ∗ Z. By the comments immediately after the statement of Theorem 6.1.1 we can assume that V at least one letter from A and at least two letters from B as a reduced word in A ∗ B. If V contains exactly two letters from B, then from the assumption that hA, V i is not isomorphic to A ∗ Z it follows that the two letters must be inverses of each other. In such case w belongs to the normal closure of A ∗ C, so G is non-trivial. Hence V contains exactly three B letters. Hence V contains exactly two letters from A and exactly three letters from B since `(V ) ≤ 5. Again it follows from the assumption that hA, V i is not isomorphic to A∗Z that V is conjugate in A∗B to a letter β ∈ B with order r < ∞. Define H to be the group
A ∗ C H := r (6.3) N((c1αc2) )
∗ ∗ =A hαi T hγi=hc2c1i C, (6.4)
p q r where T = hα, γ | α , γ , (αγ) i and p, q are the orders of α and c2c1 respectively.
Then G = (A ∗ B)hA,V i ∗hA,c1αc2i H, provided of course that hA, c1αc2i embeds in H. We show below that this is in fact the case. By Bass-Serre theory, H acts on a tree Γ. Let the vertex set of Γ be X. There exists an edge e divides Γ (see Figure 6.1) into two components Γ1 and Γ2 with vertex sets X1 (containing vertices uA, uT ) and and X2 (containing vertices uC , uc1(uT )) respectively such that X = X1 ∪ X2. The vertices uA, uT and uC have stabilizers
A, T and C respectively. Similarly, hαi and hγi are the stabilizers of e1 and e −1 respectively. Let e2 = c1(e), then the stabilizer of e2 is c1hγic1 .
Figure 6.1: Diagram showing a section of the tree Γ on which H acts.
We aim to use the Ping-Pong Lemma to show that the subgroup of H generated by
A and the element c1αc2 is the free product A ∗ hc1αc2i. Since
A ∩ hγi = hαi ∩ hγi, (6.5)
it follows that for any non-trivial element a ∈ A, a(X2) ⊂ X1. Similarly since αγ
114 Chapter 6: One-relator product of three non-trivial groups with short relator
−1 stabilizes an edge e3 incident to uT other than e, c1αc2 = c1αγc1 stabilizes an edge e4 = c1(e3) incident to uc1(uT ) other than c1(e) = e2. If we can show that hγi and hc1αc2i intersect trivially, then it will follow that b(X1) ⊂ X2 for every non-trivial element b ∈ hc1αc2i.
−1 But hγi ∩ hc1αc2i stabilizes both e and e4, and hence also e2. So c1hγic1 is contains −1 hγi ∩ hc1αc2i. Hence hγi ∩ hc1αc2i = 1 in T implies that in c1T c1 ,
−1 −1 c1hγic1 ∩ c1hαγic1 = 1 (6.6)
Thus we have that hγi ∩ hc1αc2i = 1 as required. Hence the Ping-Pong Lemma yields the result.
It follows in particular from Lemma 6.2.1 that `(U) ≥ 2. The rest of the arguments we present rely heavily on Nielsen transformations. We transform {U, V } into a more suitable Nielsen equivalent set depending on the subgroup they generate.
Lemma 6.2.2. Suppose hU, V i is free, then G is non-trivial.
Proof. First we suppose hU, V i is free of rank 1 say with generator t. Then w can r s s r be expressed in the form w = c1t c2t , where V = t and U = t for integers s, r. If s + r = 0, then G 6= 1 since w ∈ N(C). Otherwise w = 1 is a non-singular equation over C. We assume that t 6∈ N(A) ∪ N(B) for otherwise G 6= 1 by Lemma 6.2.1. It follows that any cyclically reduced conjugate of t has length at least 2. So since s, r 6= 0 and `(tn) ≥ 2n for any n,
|s| + |r| ≤ 3. (6.7)
Consider the group (C ∗ hti) H := . N(w) If s, r ≥ 1, then C embeds in H by [67]. Otherwise without loss of generality s = −1 and r = 2. Again C embeds in H by [52]. If t is trivial in H, the H = C and so w ∈ N(A ∗ B); again G 6= 1. If t has finite order m > 1 in H, then
(A ∗ B) G = ∗ H. N(tm) hti
By the previous comment it follows that `(t) = 2–say t = αβ, so (A ∗ B)/N(tm) is a triangle group and t has order m as required. Finally if t has infinite order in H, then
G = (A ∗ B) ∗hti H.
115 Chapter 6: One-relator product of three non-trivial groups with short relator
Hence G is non-trivial. Now suppose hU, V i is free of rank 2. Let H := (C ∗ hU, V i)/N(w) = C ∗ hUi.
Note that the subgroup of H generated by {U, c1Uc2} is a free group of rank 2. It follows that G is the free product of H and A ∗ B amalgamated over the subgroups hU, c1Uc2i and hU, V i. It follows that G is non-trivial.
Lemma 6.2.3. Suppose hU, V i is isomorphic to Cp ∗ Cq or Cp ∗ Z, where Cp and
Cq are finite cyclic groups. Suppose further that `(V ) < 4. Then G is non-trivial.
By assumption hU, V i has an element of finite order. So Nielsen transformations can be applied to {U, V } to get a new set {u, v} with u or v having finite order.
Note also that V 6= α1β1α2β2. Lemma 6.2.3 is a corollary to Propositions 6.2.4–6.2.7 below.
±1 Proposition 6.2.4. If U = (αβ) and V = α1β1, then G is non-trivial.
Proof. First suppose U = (αβ)−1. Since UV is not in the normal closure of A or B, neither α = α1 nor β = β1 holds. Hence hU, V i is free of rank 2. The result follows from Lemma 6.2.2.
Suppose then that U = αβ. If α 6= α1 and β 6= β1, then {U, V } is Nielsen reduced, so hU, V i is free of rank 2 and the result follows from Lemma 6.2.2. Hence we may assume without loss of generality that α = α1. If β = β1, then hU, V i is isomorphic to Z. In this case the result follows from Lemma 6.2.2. ˆ ˆ ˆ ˆ If c1 = c2 = c then we can replace w with UβUβ1, where U = cα. In this case hUi is free of rank 1 and the result follows from Lemma 6.2.2.
Hence we may assume that β 6= β1 and c1 6= c2, so hβc2, β1c1i is free of rank 2. Again we apply Lemma 6.2.2 to show that G is non-trivial.
±1 Proposition 6.2.5. If U = (αβ) and V = α1β1α2, then G is non-trivial.
−1 Proof. Suppose U = αβ. We can assume that β 6= β1 as otherwise G maps onto −1 B, hence non-trivial. Also α1 6= α2 by Lemma 6.2.1. Since α1α2 6= 1, either α = α1 −1 −1 and β = β1 or α = α2 and β = β1 , as otherwise hU, V i is free. Hence without loss of generality we assume β = β1 and α = α1. If c1 = c2 = c, then G surjects 2 0 0 onto (B ∗ C)/N((βc) ), so non-trivial. Otherwise take U = c2α and V = α2c1α, and so hU 0,V 0i is free. Hence G is non-trivial by Lemma 6.2.2. The proof for the case where U = (αβ)−1 is similar by symmetry.
Proposition 6.2.6. If U = αβα1 and V = α2β1α3, then G is non-trivial.
Proof. We may assume that β 6= β1, as otherwise G surjects onto B. Without loss of generality, there are two possibilities to consider. Either α = α2 and β = β1 or
116 Chapter 6: One-relator product of three non-trivial groups with short relator
−1 −1 α = α2 and α1 = α3. (Note that we can not have α = α3 and α1 = α2 , for otherwise w is contained in the normal closure of B ∗ C).
0 0 In the first case, we take take U = α1c2α and V = α3c1α. By Lemma 6.2.2, we can 0 0 assume that hU ,V i is not free. Hence either c1 = c2 = c or α1 = α3. In either case G maps onto
(B ∗ C) (A ∗ B) 2 or 2 N((cβ) N((αβα1) ) respectively. Hence G is non-trivial.
In the second case where α = α2 and α1 = α3, we can replace B with its conjugate by α, and w by W = c1βαc˜ 2β1α˜, whereα ˜ = αα1. Since G is isomorphic to
(A ∗ α−1Bα ∗ C) G0 = , (6.8) N(W ) the result follows from Lemma 6.2.4.
Proposition 6.2.7. If U = αβα1 and V = β1α2β2, then G is non-trivial.
−1 −1 Proof. In this case, the only possibility is α = α1 or β1 = β2 . Either case, the result follows from Lemma 6.2.1.
±1 Finally, we consider the case where U = (αβ) and V = α1β1α2β2. Since hU, V i =
Cp ∗ K, without loss of generality we have by Nielsen transformations that either
1. U = αβ and V ∈ {αβαβ2, αβα2β} with β 6= β2, or
−1 −1 2. U = β α and V ∈ {αβαβ2, αβα2β} with α 6= α2.
Remark 6.2.8. In (1) and (2) above we gave two forms of V . If V = αβα2β we can replace U and V by βUβ−1 and βV β−1 respectively (or equivalently replace C by −1 β Cβ) and interchange A and B to get the first αβαβ2. In what follows we regard G as a one-relator product of A∗B and C. For convenience −1 −1 −1 we let U1 = αβ and U2 = β α , so U2 = U1 . We use R to denote a relator in G which is a cyclically reduced word in {U, V } and `(R) denotes its length also as a word in {U, V } .
Definition 6.2.9. The index of R is the number of cyclic subwords of the form (UU)±1,(VV )±1,(VU −1)±1 or (U −1V )±1.
Definition 6.2.9 generalizes the notion of sign-index. Recall that the sign-index of R is n (necessarily even) if a cyclic permutation of R has the form
−1 −1 W1W2 W3 ...Wn−1Wn ,
117 Chapter 6: One-relator product of three non-trivial groups with short relator
with each Wi a positive word in {U, V }. In particular the index of R is bounded below by its sign-index, and above by `(R). By Remark 6.2.8
±1 Nielsen transformation −1 {U, V } = {(αβ) , αβαβ2} −−−−−−−−−−−−→{αβ, β β2}.
−1 There are two possibilities to consider. If {αβ, β β2} is Nielsen reduced, then −1 −1 hU, V i is free (if β β2 has infinite order), or Z ∗ Zm (if β β2 has order m). A −1 second possibility is that {αβ, β β2} is not Nielsen reduced. In which case β is a −1 −1 power of β β2, so B is cyclic (generated by β β2). In particular it follows that −1 β β2 must have order at least 3 (since by assumption β 6= β2).
Proposition 6.2.10. Suppose that R is a cyclically reduced word in {U, V } and that −1 2 (β β2) 6= 1. Then `(R) ≥ 2(6 − k), where k is the index of R.
Proof. It is easy to see that (UV )m 6= 1. Hence since also k ≤ `(R), we can assume the 1 ≤ k ≤ 3. In particular it is enough to consider the cases when R has sign-index 0 or 2.
First if U = U1, then U and V generates a free sub-semigroup of A ∗ B of rank −1 2 2 (since (β β2) 6= 1). In which case R is a positive word which is trivial in a free sub-semigroup or we get an equality between two positive words in a free sub- semigroup, depending on whether the sign-index is 0 or 2 respectively. None of these can happen.
Suppose then that U = U2 and `(R) < 2(6 − k). Up to inversion and cyclic permu- tation the list of such R is contained in Table 6.1 below. It is easy to check that none of the words appearing in the table is trivial.
−1 2 Remark 6.2.11. We remark that if (β β2) = 1, then B can not be cyclic since that −1 will imply that β = β2. In other words {U, β β2} is Nielsen reduced.
118 Chapter 6: One-relator product of three non-trivial groups with short relator
Sign-index Index Length List 0 1 < 10 U 2VUV U 2(VU)2V U 2(VU)3V V 2UVU V 2(UV )2U V 2(UV )3U 0 2 < 8 U 3VUV U 2V 2 U 2V 2UV U 2VUV 2UV V 3UVU 0 3 < 6 U 3V 2 V 3U 2 2 2 < 8 U −1(VU)2V (VU)−1(UV )2 (UVU)−1VUV V −1(UV )2U 2 3 < 6 U 2(VUV )−1 U 2V (VU)−1 U 2VU −1V U 2V −1UV V 2(UVU)−1 V 2U(UV )−1 V 2UVU −1 V 2UV −1U V 2U −1VU V 2U −1V −1U
Table 6.1: A table containing all the possible words R (up to cyclic permutation and inversion) in {U2,V } of index k, 1 ≤ k ≤ 3 and `(R) ≥ 2(6 − k).
119 Chapter 6: One-relator product of three non-trivial groups with short relator
±1 −1 Lemma 6.2.12. If U = (αβ) , V = α1β1α2β2 and {U, β β2} is not Nielsen reduced, then G is non-trivial.
Proof. Suppose by contradiction that G is trivial, then we get a non-trivial W- minimal spherical picture M over G with an exceptional region, where W is the set of non-trivial elements in A ∗ B ∪ C.
−1 Note that both A = hαi and B = hβ β2i are cyclic. So by Theorem 4.6.1 we assume that C is neither cyclic nor dihedral. Hence the only relations between c1, c2 2 of length less than 4 are powers of c1 or c2. Without loss of generality, c1 = 1. But 4 then the only possible relator of length 4 involving c2 is c2, and so c2 has order 3 or 4 in C. By bridge-moves, we can assume that no two adjacent corners in a C-region are labelled by c1 unless it is a 2-gon. Similarly any C-region has at most 2 or 3 consecutive c2-corners depending on whether c2 has order 3 or 4 respectively.
Every c1-corner gets angle 0, every c2-corner gets angle π/3, and every U- and V - corner gets angle 5π/6. This ensures that vertices have curvature 0, and C-regions have non-positive curvature. However, (A ∗ B)-regions can have positive curvature. We overcome this by redistributing any such positive curvature to neighbouring negatively curved C-regions, as follows. Let R be the label of an interior region of ∆ of M. If R has index n, we transfer
π/3 of curvature across each of the edges of ∆ joining a c1-corner to c2-corner of an adjacent C-region.
Figure 6.2: Diagram showing an (A ∗ B)-region with label (U 2V )2UV of index 2. The head of the red arrows indicate the adjacent C-regions receiving π/3 curvature.
n Now any interior (A∗B)-region whose label is of the form (U2U2V ) (up to inversion) has index n and curvature at most π/2. However, n > 2 by Remark 6.2.11. Hence it
120 Chapter 6: One-relator product of three non-trivial groups with short relator has transferred at least π of curvature to neighbours, so becomes negatively curved. Similarly, it follows from Proposition 6.2.10 that any interior (A ∗ B)-region whose n label is of the form (U2U2V ) (up to inversion) has curvature at most π, and has become non-positively curved after transfer as well. It follows that interior (A ∗ B)- regions non-positively curved after transfer.
A C-region ∆ receives π/3 of positive curvature across each edge separating a c1- corner from a c2-corner. Suppose that in ∆ there are p c1-corners, q c2-corners, and r edges separating a c1-corner from a c2-corner. The curvature of ∆ after transfer is at most
qπ rπ (6 − 2p − q)π 2π − (p + q)π + + ≤ . (6.9) 3 3 3 If 2p + q ≥ 6, then ∆ still has non-positive curvature after transfer. Suppose ∆ is an interior C-region and 2p + q ≤ 5. Then either p = 0 or q = 0 (since p, q 6= 1). Hence curvature after transfer is (2 − p)π or (6 − 2q)π/3, both of which are at most 0 since p ≥ 2 and q ≥ 3. Since the curvature of the exceptional region is less than 4π, we get a contradiction that curvature of M is 4π. Hence G non-trivial.
±1 −1 Lemma 6.2.13. If U = (αβ) , V = α1β1α2β2 and {U, β β2} is Nielsen reduced, then G is non-trivial.
Figure 6.3: Positively oriented vertices of Γ when n = 2. The figure on the left cor- 2 n −2 n responds to a vertex of Γ when r = (t c1tc2) and the other is when r = (t c1tc2) .
−1 Proof. By assumption hU, V i = hUi ∗ hβ β2i, and is isomorphic to Z ∗ Zn, where −1 n > 1 is the order of β β2. We can assume by Theorem 4.6.1 that hc1, c2i is not dihedral. Consider the relative presentation
±2 n H = hC, t | r = (t c1tc2) i.
121 Chapter 6: One-relator product of three non-trivial groups with short relator
The aim is to show that G is the free product of H and A ∗ B amalgamated over the subgroups hU, V i and ht, c1tc2i. To this end we must show that the latter is also isomorphic to Z ∗ Zn. In other words, any relation in H which is a word in {t, c1tc2} ±2 n is a consequence of (t c1tc2) . If this is not so, we obtain a W-minimal non-trivial picture Γ over H on D2, where
W is the set of non-trivial elements in the free product hti∗hc1tc2i. Figure 6.3 shows typical vertices with positive orientation in the case of n = 2. Note that there are no 2-zones with corners labelled by 1±1, for in such case, either the two vertices cancel, or we can combine the vertex with the boundary. In both cases we get a smaller picture, thereby contradicting minimality of Γ. Make regions of Γ flat by assigning angle (d(∆)−2)π/d(∆) to each corner of a region ∆ of Γ of degree d(∆). We claim that interior vertices Γ are non-positively curved. 2 n −2 n The proof is in two stages depending on whether r = (t c1tc2) or r = (t c1tc2) .
2 n Suppose that r = (t c1tc2) . Since r is a positive word, Γ is bipartite. More precisely, only vertices of opposite orientations are adjacent in Γ. In particular this implies that regions have even degrees. Every interior vertex v bounds at least two regions with a corner labelled 1. By minimality of Γ, every such region has degree at least
4. Also every 2-zone gives the relation c1 = c2, and so each of the two regions on both sides of the 2-zone has a corner labelled 1, hence is at least a 4-gon. Hence v bounds at least four regions of degree at least 4 and so is non-positively curved.
−2 n The case of (t c1tc2) is slightly different as regions can be odd. Note that any corner is either a source (the two arrows point outwards), sink (the two arrows point inwards), or saddle (one arrow points inwards and the other points outwards) depending on whether it is a c1-, c2- or 1-corner (see Figure 6.3). So in particular 2 2 any 2-zone gives the relation c1 = 1 or c2 = 1. If an interior vertex v does not bound a 2-zone, then v satisfies C(3n). Suppose it does. As before any region adjacent to a 2-zone has a 1-corner. It follows that such a region has degree at least 4. There are at least two regions with 1-corner at v. If v bounds only one 2-zone, then it has degree at least 5 and bounds at least three regions of degree 4. Otherwise v bounds at least four 4-gons. In all cases v has non-positive curvature. It follows that there exists a boundary vertex of degree at most 3. This is clearly impossible if n > 2 (since we will get a 2-zone with corners labelled 1). So we assume that n = 2. An argument similar to the ones given above shows that such a vertex 2 must connect to ∂D by an ω-zone, with ω ≥ 3. It follows that either one of c1 or c2 is trivial or we can combine such a vertex with ∂D2 to get a smaller picture. Both possibilities lead to a contradiction which completes the proof.
122 Chapter 6: One-relator product of three non-trivial groups with short relator
6.3 Main result
Let us recall that the statement of the theorem: if A, B and C are non-trivial groups, then the one-relator product G = (A∗B∗C)/N(w) is non-trivial where w ∈ A∗B∗C is a cyclically reduced word with `(w) < 9.
Proof of Theorem 6.1.1. By the comments immediately after the statement of The- orem 6.1.1, can assume 6 ≤ `(w) < 9 and w has the form w = c1Uc2V (up to cyclic permutation) where U, V ∈ A ∗ B and c1, c2 ∈ C with c1c2 6= 1. It follows from Grushko’s theorem that the subgroup of A ∗ B generated by U and V is isomorphic to one of the following:
1. Conjugate to subgroup of A (or B).
2. Free group of rank one.
3. Free group of rank two.
4. Free product of two finite cyclic groups.
5. Free product of finite and infinite cyclic groups.
In the case of part 1 the result follows from Lemma 6.2.1. For parts 2 and 3, the result follows from Lemma 6.2.2. And finally the result follows from Lemmas 6.2.3, 6.2.12 and 6.2.13 in the case of parts 4 and 5. This completes the proof.
123 Chapter 7
Conclusion and future work
In this work we proved a number of results about one-relator groups under some conditions. In particular we produced various lower bounds on the length of the relator of a one-relator product G for which G can be trivial (in the case of three factors) or fails to satisfy the Freiheitssatz (in the case of two factors). We now mention some future work.
7.0.1 Removing admissibility condition
In Chapter 3 we proved various results about one-relator products induced from generalised triangle groups. This was not without the requirement of admissibility (as in [62]) or that the relator had no element of order 2. Given also the conjecture of Freiheitssatz for any one-relator products in which the relator is a proper power, it seems possible that most, if not all these results are still true without these requirements. Hence we make the following conjecture.
Conjecture 7.0.1 (Conjecture). Let G be a one-relator product induced from a generalised triangle group H with factor groups G1 and G2. Let H be the quotient of hai ∗ hUbU −1i by N(Rn). Then
1. The maps G1 → G, G2 → G and H → G are all injective.
2. If the word problems are soluble for H, G1 and G2, then it is soluble for G.
n 3. If some cyclic permutation of R has the form W1W2 with `(W1) > 0 and
`(W2) > 0 as words in G, then W1 6= 1 6= W2 as words in G. In particular R has order n in G.
A special case of Conjecture 7.0.1 which surfaced in at least two places in this work is the case where the generalised triangle group H has the form
H = hx, y | x2, yp, w(x, y)2i, (7.1)
124 Chapter 7: Conclusion and future work for some integer p > 2. Perhaps one can try to deal with this special case.
7.0.2 Solution of equations over free product of groups
For some non-trivial groups A and B, and the infinite cyclic group hti, consider the one-relator product of groups (A ∗ B) G = (7.2) N(w) for some word w ∈ A ∗ B with `(w) ≥ 2. For the rest of this chapter w(t) ∈ A ∗ B ∗ hti is a word obtained from w ∈ A ∗ B by inserting powers of t between the letters in w. For example suppose w = a1b1 . . . anbn
α1 α2 α2n−1 α2n with ai ∈ A and bi ∈ B. Then w(t) = a1t b1t . . . ant bnt for integers αi, i = 1,..., 2n. We are interested in the relationship between Freiheitssatz for G and solution of the equation w(t) = 1 over A ∗ B.
Theorem 7.0.2. Suppose w(t) is as above such that |αi| = 1 and αi = −αi+1. Then G satisfies the Freiheitssatz if and only if w(t) = 1 has a solution over A ∗ B.
Proof. By assumption and without loss of generality, we can write w(t) in the form
k Y −1 w(t) = aitbit , i=1 Qk where w = i=1 aibi. Let K be normal closure of A ∗ B in the universal solution group A ∗ B ∗ hti H = . N(w(t)) Then
∗ (tkAt−k ∗ tkBt−k) K = k∈Z N({tkw(t)t−k}; k ∈ Z) However since tkw(t)t−k is contained tkAt−k ∗ t(k+1)Bt−(k+1) for each k, it follows that K = ∗k∈ZGk, where
(tkAt−k ∗ t(k+1)Bt−(k+1)) G = . k N(tkw(t)t−k)
If G satisfies the Freiheitssatz, then A embeds in G0 and B embeds in G−1. This implies that A ∗ B embeds in H (via K). On the other hand suppose w(t) = 1 has a solution over A ∗ B. Then α ∈ Ker(A → G0) ⊆ Ker(A → K) ⊆ Ker(A ∗ B → H) implies α = 1. Similarly if we replace A by B and G0 by G−1. This implies that Freiheitssatz holds for G.
125 Chapter 7: Conclusion and future work
Perhaps one can ask the following more general question.
Question 7.0.3. Is it true that Freihetssatz holds for G = (A ∗ B)/N(w) where w = a1b1 . . . anbn, ai ∈ A and bi ∈ B, whenever the equation
α1 α2 α2n−1 α2n w(t) = a1t b1t . . . ant bnt = 1 has a solution over A ∗ B?
As mentioned in Chapter 2, Question 7.0.3 has a positive answer when A and B are locally indicable groups, and is conjectured to be true in the more general setting where both are torsion-free. From Theorem 7.0.2 one might suspect that not only does Question 7.0.3 have a positive answer in general, but also the reverse direction also has a positive answer. In other words, Freihetssatz holds for G if and only if w(t) = 1 has a solution over A ∗ B.
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